MCB 421 First Exam October 4, 2004 1. (10 pts). An E. coli strain (strain A) that lacks an inducing prophage and carries the F factor is heavily irradiated with UV light and then mixed 1:1 with a second E. coli strain (strain B) that carries a wild type lambda phage as a prophage and lacks the F factor. What would be the observed outcome of this experiment? Would the outcome differ if the second strain contained a deletion of the recA gene (∆recA)? What would happen if strain B carried the F factor in addition to the lambda prophage?

Conjugational transfer of the UV-damaged F factor into the F- strain will induce the SOS operon and hence will induce the lambda prophage. (This experiment resembles the Weigle mutagenesis experiments discussed in class except that a different assay is used for SOS induction and the damaged DNA enters by a different process.) If an F+ strain is the recipient no mating will occur (due to a lack of interaction between the donor and recipient cells called surface exclusion) so no induction of lambda will be seen. (This result is real)

2. (40 pts) The enzyme, DNA gyrase, is essential for DNA replication because it relieves the torsional stress that results from separation of the DNA strands during replication. Gyrase is composed of two protein subunits (A & B) encoded by the gyrA and gyrB genes. Gyrase is the target of many widely used antibiotics (e.g., Cipro), some of which inhibit gyrase by binding to the A subunit whereas others bind to the B subunit. The paradigm A subunit drug is naladixic acid (nal) and mutants resistant to high concentrations of nal (nalR) map almost always in the gyrA gene.

a) Mutants resistant to high concentrations of nal are found that map in the gyrB gene. These mutants are very rare. Give a brief molecular explanation.

The mutation alters GyrB such that upon interaction with GyrA, GyrA can no longer bind nal. These mutations are very rare since they must alter the GyrA-GyrB interface in a way that is communicated to the nal binding site of GyrA. (Another real result). MCB 421 First Exam October 4, 2004

b) When wild type strains of E. coli are treated with high concentrations of nal, the DNA becomes fragmented. This is because the A subunit cuts both strands of the DNA. Neighboring regions of intact DNA are then passed thought the break thereby relieving the torsional stress and the A subunit then restores the phosphodiester bonds to give intact DNA lacking the torsional stress. Nal inhibits the resealing of the DNA and thus the DNA becomes degraded. What do you think would happen if a high concentration of nal was added to a wild type E. coli strain that carries a wild type lambda prophage? Why?

The fragmented DNA induces the SOS system and hence lambda is induced. Since gyrase gives double stranded breaks and SOS induction requires single stranded DNA fragments, some other enzyme must convert the double stranded fragments into single stranded DNA fragments. (This is mainly the RecBCD nuclease). (Another real result).

c) Given the above data would you expect nalR to be dominant or recessive to the wild type allele? That is, would a strain carrying one copy of a nalR gyrA gene on a F’ and a wild type gyrA gene on the chromosome be sensitive or resistant to nal? (Think carefully about this one and justify your answer).

The strain carrying one WT copy of gyrA and one nalR copy of gyrA is sensitive to nal. That is, nalS is dominant over nalR. This seems perverse, but remember that the cell will contain both kinds of gyrase. When nal is added the sensitive gyrase will fragment the DNA which will kill the cells. (Another real result).

d) There is class of mutants resistant to only low concentrations of nal that map neither in gyrA or gyrB. Give a brief explanation for these mutants.

These mutations turn on a membrane pump that pumps the nal out of the cell. The pump is low capacity and thus it gives resistance to only low concentrations of nal. (A real result discussed in class). MCB 421 First Exam October 4, 2004 e) Explain how you might use the above information about gyrase to determine the time course of chromosomal gene transfer after mixing an Hfr donor strain and an F- recipient strain. Note that at high concentrations nal acts almost instantaneously.

Use a nalR recipient strain and a nalS donor strain. Transfer of DNA during mating requires DNA synthesis, so adding nal will stop transfer immediately. This works as well as hydrodynamic shearing and is a lot easier. The nalR also provides a counter-selection marker. MCB 421 First Exam October 4, 2004 3. (10 pts) A mutant strain of Salmonella was isolated that grows as well as the wild-type parent on rich medium, but when transferred from rich medium to minimal medium the strain shows a very long lag phase compared to the wild-type parent. In the figure below, growth of the wild type is shown by squares and the mutant strain is shown by triangles.

How could you determine if the growth that occurred after the long lag phase was due to a second mutation that permitted growth in minimal medium or simply due to a requirement for longer time for adaptation to the minimal medium?

If there is a second mutation then cells taken from the fully grown mutant (∆) minimal medium culture should grow as soon (and as well) as the wild type upon shift from rich medium to minimal medium. That is, the character should be inherited.

If the result is due to adaptation, then the cells taken from the fully grown mutant minimal medium culture will grow with the same lag as seen before. That is, the character will not be inherited. MCB 421 First Exam October 4, 2004

4. (15 pts) A mutation in the dnaQ gene (called mutD5) which encodes the 3’-5’ exonuclease of E. coli DNA polymerase III gives a 104 to 105 increase over the normal rate of mutation when a mutD5 strain is grown on a rich medium (cell doubling time of 25 min). However, when the same strain is grown in a minimal medium (cell doubling time of 120 min) the increase in mutation rate drops to only a 250-fold increase.

If a multicopy plasmid (30 copies/cell) carrying a wild type mutH gene is introduced into the mutD5 strain, the mutation rate in rich medium (cell doubling time of 25 min) is a 250-fold increase over the normal rate.

Give a molecular explanation for these data that includes explanations for the effects of growth medium and the plasmid.

Deductions are:

i) The first paragraph says that rate of mutagenesis depends on the growth rate. Since you know that the rate of DNA synthesis depends on the growth rate, it follows that the rate of mutagenesis depends on the rate of DNA synthesis. (In addition you were told that DnaQ is a subunit of Pol III).

ii) You were given the fact that DnaQ is responsible for editing out mistakes made by Pol III. As you learned most of the remaining errors are fixed by mismatch repair. In the mutD5 strain DnaQ function is absent or weak so that mutations arise at very high frequency. When the cells are growing slowly (hence making DNA slowly) on minimal medium, mismatch repair can cope with most of the mutations and mutagenesis increases only 250-fold.

iii) However, when the cells are growing rapidly (hence making DNA rapidly) mismatch repair cannot keep up and the mutation rate goes way up (to 104-105). Therefore, mismatch repair is titrated. This is clear from the second paragraph where overproducing MutH removes the effect of growth rate (hence MutH must be the limiting mismatch repair component).

This is a real effect. The mutD5 strains are often used to mutagenize plasmids and phages. The remaining mutagenesis is due to the fact that mismatch repair cannot fix C-C mismatches and is poor at fixing some other mismatches. MCB 421 First Exam October 4, 2004 5. (15 pts) Phage 186 of E. coli is a relative of phage lambda, but has important differences. A major difference is that phage 186 makes an antirepressor protein that binds to and prevents DNA binding by the phage 186 cI repressor. Consider the following facts:

a. Phage 186 forms turbid plaques on a wild type E. coli strain, forms clear plaques on a ∆lexA (lexA deletion) strain. Phage lambda forms turbid plaques on both wild type and ∆lexA E. coli strains. b. Phage 186 lysogens of a lexA(ind-) strain are not induced by UV irradiation. c. Phage 186 lysogens of a lexA (Ts) host are induced upon shift to 42oC. d. Phage 186 lysogens of a lexA (Ts) ∆recA host are induced upon shift to 42oC.

Compare and contrast the induction of phages 186 and lambda and provide a molecular mechanism that explains the induction of phage 186 under the above conditions.

The answer is that the promoter responsible for transcription of the phage 186 gene that encodes the anti-repressor protein (which is called Tum) is directly controlled by LexA. That is LexA directly represses Tum expression. Hence:

a. Phage 186 makes a clear plaque on ∆lexA strain. Remember that clear plaques means no lysogens were formed. That is, all of the phage went the lytic route. This is because there is no LexA in these cells to shut off Tum synthesis, so lots of Tum accumulates which binds to the phage cI repressor and keeps it from shutting down the genes of lytic growth. Hence, 186 phage cannot lysogenize.

b. There will be no clipping of LexA in this strain upon UV irradiation so the Tum promoter will remain off so no induction will occur (as also seem in lambda).

c. Mutant LexA is inactive at high temp so Tum is expressed and lots of Tum accumulates which binds to the phage cI repressor and neutralizes it. Since there is active cI repressor phage 186 is induced.

d. This shows that RecA plays no direct role in 186 induction, it only works through LexA. Hence, RecA cannot be not a coprotease required for clipping 186 cI repressor.

6. (10 pts). A virus is isolated that has a genome consisting of a single DNA molecule having the following base composition: A = 24.6%, T = 34.7%, G = 20.2%, C = 20.2%. Briefly describe the DNA molecule and how you would determine if the molecule is linear or circular.

A does not equal T so the DNA cannot be double stranded. It must be single stranded. The easiest test of circular versus linear DNA is to add a DNA exonuclease and test for hydrolysis of the DNA. For example Pol I in the absence of dNTPs. Exonucleases need a DNA end, no ends in circular DNA so no hydrolysis. This is phage phiX174 which has a circular single stranded genome of this base composition.