In: Biology
An E. Coli strain has the following lac genotype: Lacl+ lacP+ lacOc lacZ- / lacl- lacP+ lacO + lacZ + If lactose is PRESENT, what would the level of Beta-galactose be?
1- High
2- Low
The answer will be high (option 1).
Explanation:
Notation of the given symbol:
I+ is normal repressor, which is
active in absence of lactose & inactive in presence of lactose,
I- is mutant repressor & it is inactive in both presence &
absence of lactose, P+ is normal promoter, O+ is normal operator;
repressor can bind with it in absence of lactose & unable to
bind in presence of lactose, Oc mutant operator; repressor unable
to bind in any condition, Z+ produce , Z-
unable to produce
.
In this strain, first allele lack
gene for enzyme
(Z-). As a result, this allele will unable to produce
in any
condition. Now, repressor protein can act in trans configuration.
As a result, repressor protein of first allele can bind with the
operator of second allele & as a result, this allele will act
as normal operon. Now, as lactose is present, repressor protein
will unable to bind with the operator of the second allele. As a
result, RNA polymerase can bind with the promoter of second allele
& transcribe the operon. Thus, high level of transcription
occurs & beta-galactose production will be high.
Note: In lac operon both glucose & lactose plays an important part for the level of transcription. If glucose is present, even in presence of lactose transcription will be low due to the fact that CAP protein will unable to bind with the DNA which helps to properly bind the RNA polymerase to the promoter. As in the given case there is no mention of glucose, we ignore the conditions regarding glucose.