BIOLOGY 1A03 Lecture Notes - Lecture 5: Beta-Galactosidase, Lac Operon, Neutral Red

Are my answers correct?

1. If a bacterium with a wildtype lac operon is grown in the presence of lactose, high levels of transcription result for a while, however, eventually transcription is again down-regulated. If, on the other hand, the sugar IPTG is present (instead of lactose), transcription levels are high and remain high. What function do IPTG and lactose share? How are they differently affected by the operon product?

Lactose and IPTG are both inducers of the Lac-Operon. Lactose is a substance that the enzyme will react with (substrate), which is split or cleaved by the Beta-galactosidase. When you add lactose expressions level of lac operon will increase but as lactose is used by the beta-galactosidase expression levels of lac operon will decrease.

IPTG is a chemical cannot be split or cleaved by beta-galactosidase. The expression level of lac operon will remain the same.

Or Lactose can be easily broken down but IPTG cannot be broken down. What does expressions level mean?

2. If a lacI - bacterium is grown in the presence of the sugar X-gal, a blue dye appears; however, the concentration of X-gal decreases over time. What has happened to the X-gal?

If a lacI - bacterium is grown in the presence of the sugar X-gal, the Lacl is mutant or the gene is damaged which alters the genetic message carried by that gene. The LacI gene produces the repressor protein which switches off the operon, but since it is damaged no repressor protein is produced and the operon is switched on. The Lacl is transcribed into beta-galactosidase (lactose) The enzyme beta-galactosidase (lactose) splits or separates the X-gal which will result in decreased concentration of X-gal with time.

3. If a wildtype bacterium is grown in the presence of the sugar X-gal, transcription of the polycistronic m-RNA does not occur. What is X-gal unable to do? Without adding lactose (or genetically modifying the bacteria), how could you fix this problem

X-gal cannot form lactose, the repressor remains off and beta-galactosidase is not formed. This can be fixed by adding IPTG which can switch on the operon

In the following points, I have summarized Lac operon in Ecoli. I need to check if this is on the right track? and where does 'cleavage of disaccharides' fit into this picture?

Lactose is a disaccharide and the bacteria need specific genes to manage catabolism

1) The lacl promotor gene synthesis the repressor protein which binds to the operator gene stopping the RNA polymerase from coding the lac operon genes

2)But when lactose is present, it needs the specific genes to catabolise lactose

3) So, When allolactose (a derivative of lactose)is present, it binds to the repressor protein, the repressor releases from the operator gene

4) This allows the RNA polymerase can attach to the DNA control region

5) And drive down the operon for transcription and induction of the enzyme lacZ(beta-galactosidase) lacY and lacA which breaks lactose down into glucose and galactose

6) When the lactose is all gone, the repressor returns to the operator in a feedback loop system

a) If a lacI - bacterium is grown in the presence of the sugar X-gal, a blue dye appears; however, the concentration of X-gal decreases over time. What has happened to the X-gal?

b) If a wildtype bacterium is grown in the presence of the sugar X-gal, transcription of the polycistronic m-RNA does not occur. What is X-gal unable to do? Without adding lactose (or genetically modifying the bacteria), how could you fix this problem (can you fix this problem by adding IPTG)

Can you explain the following answer in simple terms? Still do not understand what happens to the X-gal, and what x-gal is unable to do. Thank you

Answer A)

X-gal stands for 5-bromo-4-chloro-3-indolyl-β-D-galactopyranoside which has glycosidic bond between galactose and 5-bromo-4-chloro-3-hydroxyindole. It gets cleaved by beta-galactosidase translated from lac operon.

If lacI- mutant is provided with X-gal, because of mutant lacI gene, no repressor will be formed and hence the operon is swiched on.

The enzyme beta-galactosidase cleaves X-gal which results in decreased concentration of X-gal with time.

X-gal gets cleaved to form 5-bromo-4-chloro-3-hydroxyindole which then dimerizes to derivative of indigo and hence blue color appears.

Answer b

X-gal cannot form allolactose or its analogue (inducer) and so in wild type where lacI gene is normal and forms repressor. So, in absence of allolactose, repressor inducer complex is not formed and so, operon remains switched off and hence no beta-galactosidase is formed.

This problem can be solved by adding IPTG i.e. Isopropyl beta-1-thiogalactopyranoside which mimics the action of allolactose and so can switch on the operon.