Question

IV(c-d). Prokaryotic Gene Regulation Background: Several genetic regions are needed for lactose metabolism: P(promoter), O(operator), lacZ(B-galactosidase gene), lacY(permease gene), lacA(transacetylase gene) and lacl(repressor gene). Strains can be defined as having c or s copies: + refers to the functional version, - refers to a defect version, c refers to a constitutively active version and s refers to a superrepressor version Each strain may or may not have an extra-chromosomal plasmid containing the genes indicated. TA-transacetylase For each strain, determine the protein expression levels. Define as: none, low or high. Presence of glucose & Absence of glucose & presencePresence of glucose & lactose Bacterial chromosome Plasmid absence of lactose of lactose ?-gall permease | TA | ?-gall permease | TA | ?-gall permease | TA none none none none none none

Background: Several genetic regions are needed for lactose metabolism: P(promoter), O(operator), lacZ(b-galactosidase gene), lacY(permease gene), lacA(transacetylase gene) and lacI(repressor gene). Strains can be defined as having + - c or s copies: + refers to the functional version, - refers to a defect version, c refers to a constitutively active version and s refers to a superrepressor version. Each strain may or may not have an extra-chromosomal plasmid containing the genes indicated. TA=transacetylase

it would be very helpful if you could explain how you got the answer for atleast one of them . thanks

Answer 1

A lac operon is a group of different coding sequences which work as a team to digest the lactose whenever it is present in the medium. The presence of glucose will negate the cause the lac operon as glucose provides for ready energy and there is no further need to digest lactose. So lac operon is closed when glucose is present in the medium, irrespective of lactose presence. The absence of glucose triggers the lac operon.
In the first case, all the genes are present and functional. When glucose is present, the expression of b-gal, permease and TA are inhibited. So we can fill 'none'. When the glucose is absent and lactose is present the system will be on the full scale, hence we fill "high".
When the glucose and lactose are both present, the lac operon expression is largely reduced but it does occur in small levels, called basal transcription. Hence we can fill low.

When you have a plasmid, it will also help in transcription. So in the third case, the plasmid can provide p+ which is p- is chromosome. p+ is dominant and the operon would work normally as above. Sometimes the operon is mutated such the Oc is found in place of O. This means the operator is constitutively or always active. So whether lactose is present or absent, the lac operon continues to express. The I- means the repression is absent, even here the lac operon can work without any influence of glucose. This means, in I- condition, the lac operon works in the presence of glucose as well.

Note that b-gal is from Z+, permease is from Y+, and TA is from A+. The absence of these genes means an absence of the gene products.

I hope this would help you to solve the rest. Please get back in comments for any query.