Question

. A wild-type Hfr is mated to an F- strain that is io+ zy- . In the absence of inducer and glucose what should happen to β-galactosidase levels immediately after the mating (increase, decrease, or stay the same)? a. Explain your answer. b. Describe what would happen if the F- recipients were i+ oc zy- . c. In diploids, would you expect cap+ to be dominant or recessive to cap- ? Refer to lac expression and describe the phenotype of cap+ , cap- , and cap+ /cap- bacteria. d. What would be the effect on lac operon expression of a mutation in cap that caused it to bind to DNA in the absence of cyclic AMP?

Answer 1

Immediately after the mating, the HFr chromosomal strand displaces and moves through the conjugation bridge into the F- strain. Since it's a wild type HFr strain, it will be io+z+y+. As it get's transferred, it will undergo homologous recombination to yield a F strain with the same genotype, where the displaced strand will be degraded. In absence of any inducer and glucose, the beta galactosidase level in the cell will remain the same. Y is the gene for encoding permease which allows lactose(inducer) or glucose to enter the cell. In it's absence, it makes no impact on the level of rise or fall of the enzyme.

With a F- strain of i+oczy-, having a constitutive operator, the presence or absence, again wouldn't have made a difference. There would have been a perennially increased amount of beta galactosidase in the strain.

In diploids, cap+ is expected to be dominant to cap-, so that cap and cAMP binding to the site, and furthur activation of the operon can proceed. Natural selection will promote the causation of the lac operon machinery.

Cap+ bacteria- cAMP will bind to CAP protein, binding to the regulatory site in the presence of low glucose, and lac operon will be activated.

Cap- bacteria- cAMP will not be able to bind to any protein, so lac operon will not be activated in the absence of glucose. The cell wouldn't be able to metabolise lactose as the food source.

Cap+/cap- bacteria- Since it's expected to be dominant, there wouldn't be any problem in transcribing lac operon genes but on mating, there is a 50% possibility of transfer of cap- strand, which again strikes on a 50% possibility to mate and give rise to a completely cap-/cap- strain (.25)

For a mutation in cap that caused it to bind to DNA in the absence of cAMP, there would have been an expression of the lac operon even in the presence of glucose. So, after a stipulated amount of time, the glucose concentration in the cell would become massively high, with the addition of the metabolism of lactose into glucose and galactose.