7.03 Genetics: Answer Key to Problem Set 4 (2001)

Problem Set 4 Answer Key 10/26/01 1.a) F’ must be I+ P+ O+ Z+ (could possibly be I- if transduced into lacZ- strain since that strain may be lacI+ and would mask an I- on the F’) and the chromosome must be either I-d or P+ Oc Z+ (we do not know if either P, O or Z are mutated in the I-d nor do we know if the I is mutated in the case of an Oc mutation). When the F’ is transferred into a LacZ- (or Lac deleted) strain we get normal regulation of the Lac Z. This suggests that the F’ is LacI+ LacP+ LacO + and LacZ+ (or in other words that it carries a complete normal lac operon). When this F’ is in our strain the result is a constitutively active lac operon. This suggests that the chromosome must have either an I-d or an Oc since both of these would result in constitutive activity (when combined with the F’ in the case of the I-d). In the case of an I-d mutation we do not know if the chromosome is Z+ or Z- but we know that for an Oc mutation to be constitutively active the chromosome would have to be Z+. b) Here we can rule out the possibility that the chromosome has an Oc mutation because that will not allow us to get the three different Hfrs that we are seeing. We also now know that the chromosome must also have either a P-, O- or Z- mutation (this is required to get the three Hfrs). Since the F’ is P+ O+ Z+ it does not matter which is mutated on the chromosome so for simplicity I will chose Z. Now we know that the chromosome was I-d Z- and the F’ was I+ and Z+ so we can now test the possible orders. F’ I+ Z+ -d ­ Chromosome I Crossovers 1 Z 2 3 Now we can determine what the Hfr resulting from each of these crossovers would look like. I+ Z+ I-d Z­ 1: I-d Z+ I+ Z­ 2: I-d Z­ I+ Z+ 3: Now we can look at which alleles will be transferred early and determine the direction of the origin of replication on the F’. From this we can see that if transfer begins left of the origin (� facing origin in Hfr) we would get transfer of normally induced operon from Hfr 1, a constitutively active operon from Hfr 2 and an uninducible operon from Hfr 3. If transfer goes to the right of the origin (� facing origin in Hfr) we would not get an operon being transferred early that is constitutively active. Therefore we know that the map of the F’ must be (note that the orientation of the origin in the F’ is opposite that when it is in the Hfr) and the recombination events are described above: I+ Z+ 1 Problem Set 4 Answer Key 10/26/01 2. The minimum number of experiments needed to deduce this model is 6. Exp 1: Look at the phenotype of the A- mutant: Will be constitutively active. Exp 2: Look at the phenotype of the B- mutant: Will be uninducible. Exp 3: Look at A-B- double mutant: Will be uninducible, since the double mutant has the same phenotype as the B- we know that B is epistatic to A (or in other words the B gene acts after the A gene in the pathway). Exp 4: Put an A+ E+ F’ into an A-E+ cell: Will ha ve normal regulation of the operon, this tells us that Ais recessive and we can deduce that it is trans-acting (from chart in notes for lecture 18) and is a repressor. Exp 5: Put a B+ E+ F’ into a B-E+ cell: Will have normal regulation. This tells us that B is recessive. Now if we look at the chart from lecture 18 we see that just knowing that it is uninducible and recessive does not narrow it down to one type of mutation so we need one more test. Exp 6: Put B+ E+ F’ into B-E+ cell: Will have normal regulation. Then put B-E- F’ into B+ E+ cell: Will also have normal regulation. This tells us that B is also trans-acting and thus according to the chart it is an activator. 3.a) From the data given we can determine that A- is dominant and trans-acting and we are told that it is uninducible. Therefore, we can look at the chart from lecture 18 and determine that it is either a repressors or activator-d. Thus A is either a repressor or an activator. b) From the data given we can determine that B- is constitutive, recessive and trans-acting. Therefore, according to the chart it is a repressor-. Thus B is a repressor. c) Model 1: A is activator-d (prevents binding of Gene B operator) and so constitutively activates Gene B. Gene A Gene B Gene E Inducer Model 2: A is repressors and so it cannot bind the inducer and is always activating B. Gene A Gene B Gene E Inducer In both models B is a repressor and B- results in constitutive activation of Gene E. 2