MCB 421 Second Exam October 27, 2004

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1 MCB 421 Second Exam October 27, (10 pts) As discussed in class in complementation studies using F plasmids complementation can be confused with the products of homologous recombination between the chromosomal segment on the F and the chromosome, In the cross F lacz + Y - by F - lacz - Y + a) Lac + colonies formed due to complementation can generally be distinguished from Lac + colonies formed by homologous recombination. What is the distinguishing feature? Complementation is usually 100%. Each cell that gets the F grows. In contrast recombination is much less efficient; 0.1% or below. b) What mutation would you introduce into the F - strain that would ensure that all Lac + colonies formed would be the result of complementation? A reca mutation preferably a reca. The lack of RecA would block homologous recombination so all colonies would be the result of complementation.

2 2. (20 pts) You mix two E. coli strains (see below for genotypes) in a 1:1 ratio and immediately begin to take samples of the cells and place them on ice. When all samples have been taken the cells are lysed and assayed for either ß-galactosidase in (a) or the AraA enzyme activity in (b). The () means all of the genes in parentheses have been deleted. a) F laci + Z + Y + with F - lac(izy). Curve (a) results from the fact that the recipient cytoplasm has no LacI repressor. Therefore when the F lac DNA enters the lac promoter can bind RNA polymerase and express the lac genes (LacZ). However the laci gene also entered the recipient cells and is expressed. When the concentration of LacI reaches a high enough level it shuts off the promoter and ß-gal expression stops hence the plateau in ß-gal activity seen. This is called escape synthesis since the lac promoter has temporarily escaped LacI repression. (Everyone gets 5 points for this unless you didn t answer at all.) b) F arac + B + A + D + with F - ara(cbad). Curve (b) results from the fact that the recipient cytoplasm has no AraC. Since the Ara operon is positively regulated by AraC (plus arabinose) you get no AraA activity. However the arac gene also entered the recipient cells and is expressed. When the concentration of AraC is high enough it turns on the arabad promoter and AraaA activity increases until maximal. The medium would have to contain arabinose. (You might think that this experiment could not work since the donor cell would produce AraA activity and obscure what happens in the recipient cell. However, remember that the donor cells can transfer DNA even when protein synthesis is inhibited. So use a recipient strain that is resistant to a protein synthesis inhibitor and a donor that is sensitive to the inhibitor. You then do the cross in the presence of the inhibitor and add the inhibitor before you add the arabinose.)

3 c) You test another lac - strain as a recipient with the same F laci + Z + Y + as in (a), This lac - strain has no LacZ or LacY activity for. However, when you mix the cells as above and assay ß-galactosidase you get curve C. Give a molecular explanation including the lac genotype of the recipient. LacI S d) F strains transfer chromosomal markers other than those that reside on the F. However these genes are transferred at times lower efficiency than the genes on the F. Provide a mechanism for transfer of the non- F genes. State the effects that a reca mutation would have on this transfer if the mutation was in the donor strain and if the mutation was in the recipient strain. This is called chromosome mobilization and works like the F+ by F- matings discussed in class except that transfer of chromosomal genes is much more efficient (see pg 293 of book). The chromosomal segment of the F provides a large region of homology with the bacterial chromosome. Hence the F can reversibly integrate into the bacterial chromosome by a single crossover to give temporary Hfr-like strains that transfer markers on the chromosome (other than those that reside on the F ) with good (but not Hfr) efficiency. A reca mutation will block chromosome mobilization by blocking homologous recombination, although it will have no effect on transfer of the F and its chromosomal markers)

4

5 Answers to #3. his100 is non-polar since downstream genes are expressed ( = complementation) and so is probably a hisc missense mutation his101 is weakly polar since downstream genes are only weakly expressed ( = weak complementation) and so is probably a hisb nonsense mutation located at the downstream end of the hisb gene. It could also be a hisb frameshift mutation that generates a nonsense mutation. his102 is non-polar since downstream genes are expressed ( = complementation) and so is probably a hisd missense mutation that is leaky (has partial HisD function that is sufficient to give weak growth). his103 is non-polar since downstream genes are expressed ( = complementation). However, the strain lacks His H, A, and F activities and must be a hishaf deletion. The deletion must leave no naked RNA for Rho protein to bind to because the mutation is non-polar. his104 is weakly polar since downstream genes are only weakly expressed ( = weak complementation). However, the strain lacks His D, C, and B activities and must be a hisdcb deletion. The deletion must generate a frame-shift mutation that in turn generates a nonsense mutation accounting for the weak polarity seen. his105 (= also answer to b). The most straightforward mutation is a his - mutation that does not map within the operon and so is not present on the F. It could be a histidinyl-trna synthetase mutant that binds histidine poorly so that its function requires high histidine concentrations in the medium. A mutation in the histidinyltrna that is hard to charge with histidine would be a second possibility. his106 (= also answer to c). The most straightforward mutation is a mutation that inactivates the his operon promoter. Being a site on the DNA is would function only in cis and the F genes would have no effect. However, remember that the histidine operon is controlled by attenuation so anything that blocked alternatives to the formation of the transcription terminator in the leader region would result in no histidine synthesis and be cis active. So anything that blocked synthesis of the histidine-rich leader peptide (no initiation codon, no ribosome binding site, a nonsense mutation, etc.) is a possibility.

6 a. I. This one is tough. The mutation shuts down the incoming operon even when there is an O C mutation. The most straightforward mutation would be a lacz mutant that makes a lacz protein that forms inactive tetramers with the F -encoded normal LacZ protein (remember LacZ is a tetramer). So having one or two mutant LacZ monomers in a tetramer would kill the activity of the good subunits in the these tetramers. Another (not very good) possibility is a super powerful LacI S that can somehow bind an O C operator, a. II. A lacz mutation that is non-polar (since it has LacY activity) so probably a missense mutation, although a small deletion is possible. a. III. This is a classical I S mutant (acts in trans to shut down the wild type operon). a. IV. This is a lacy mutation. All you know is that it encodes a protein defective in LacY activity. b. A laci temperature sensitive mutant. The LacI is non-functional at 40 o, but functions at 30 o. c) This is a mutation in cya so no adenylate cyclase activity. Cyclic AMP (camp) is needed in order for Crp to bind and activate the lac promoter.

7 (a) Plate 1. his + Plate 2. pro + Plate 3. arg + Plate 4. pro + his + Plate 5. arg + his + Plate 6. arg + pro + (b) Pro His Arg % % <0.1%

8 Since the same Hfr was used there must be two different arg- mutants that map in different places on the chromosome (see next page). In mutant 1 the arg mutation lies between the origin of transfer and the met gene. In mutant 2 the arg mutation is on the side of the met gene opposite that of the origin of transfer.

9 arg1 met arg2 origin of transfer ~ 20 min ~ 1.5 min

10 Enzyme Levels Non-Induced Induced Genotype Wild type i + o + z + y + <0.1 < Parental mutant Revertant Type A Revertant Type B Revertant Type C i S o + z + y + <0.1 <0.1 <0.1 <0.1 i - o + z + y i S o C z + y i + o + z + y + <0.1 < The Type A mutation inactivates the LacI gene so it either makes no functional protein or no protein at all. The Type B mutation alters the operator so it no longer binds LacI. The Type C mutation reverts the laci S mutation to laci +. The strain is now wild type. 8 (10 pts) Give the definitive test that you would do in order to distinguish a positively controlled set of genes from a negatively controlled set of genes. Delete the regulatory gene. If control is positive expression of the genes will be off (non-inducible). If control is negative expression of the genes will be on (constitutive). Clues to mode of control can come from the frequencies of regulatory mutants. If non-inducible mutants are frequent and constitutive mutants are rare then the system is probably positively controlled. If non-inducible mutants are rare and constitutive mutants are frequent then the system is probably negatively controlled.

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