Question

Please explain clear and different to the answers that are already in other websites. Thank you

1. An interesting mutation in lacI results in repressors with 110-fold increased binding to both operator and nonoperator DNA. These repressors display a “reverse” induction curve, allowing β-galactosidase synthesis in the absence of an inducer (IPTG) but partly repressing β-galactosidase expression in the presence of IPTG. How can you explain this? (Note that, when IPTG binds a re- pressor, it does not completely destroy operator affinity, but rather it reduces affinity 110-fold. Additionally, as cells divide and new operators are generated by the synthesis of daughter strands, the repressor must find the new operators by searching along the DNA, rapidly binding to nonoperator sequences and dissociating from them.)

2. Certain lacI mutations eliminate operator binding by the Lac repressor but do not affect the aggregation of subunits to make a tetramer, the active form of the re- pressor. These mutations are partly dominant over wild type. Can you explain the partly dominant I− pheno- type of the I−/I+ heterodiploids?

3. You are examining the regulation of the lactose operon in the bacterium Escherichia coli. You isolate seven new inde- pendent mutant strains that lack the products of all three structural genes. You suspect that some of these muta- tions are lacIS mutations and that other mutations are al- terations that prevent the binding of RNA polymerase to the promoter region. Using whatever haploid and partial diploid genotypes that you think are necessary, describe a set of genotypes that will permit you to distinguish be- tween the lacI and lacP classes of uninducible mutations.

Answer 1

1. Given condition: laci mutation results in increased binding to operator and non operator DNA. Absence of IPTG- beta-gal synthesis occurred, Presence of IPTG – beta gal synthesis repressed.

lac operon: It has structural genes (Z,Y,A), overlapping operator and promotor region, repressor binding region where the protein formed from lacI gene binds. The repressor protein usually searches for the operator by binding to the DNA at various regions. There might be some sequence in the DNA which mimics the operator, but the repressor binds and dissociates within seconds at such sites. Thus it doesn’t take much longer for the repressor to find the actual operator.

But if the mutation has increased the affinity of repressor then it will spend more time for searching the actual operator as it might bind to sequences which mimic operator and due to increased affinity might take more seconds to dissociate from it.

Now in this time lap, we can observe the partial synthesis of beta-gal evenif IPTG is absent.

Presence of IPTG – beta gal synthesis repressed: This can be explained as follow: IPTG when binds with repressor it brings its efficiency to normal. If that is the case then repressor with normal affinity will not spend much time with regions that mimic operator but it will immediately find operator and repress the synthesis of beta-gal.

2. We are talking about heterodiploid here i.e I-/I+ heterodiploid. Lac I codes for the repressor which has four subunits. In the given case two types of repressors would be formed, one which has heterotetramer, and the other which has homotetramer. Though heterotetramer formation occurs (due to I-/I+) they do not bind to operator and are also dominant over wild type thus they will cause some expression of lac genes in the absence of lactose.

3.