Gene Regulation in Prokaryotes – Lac Operon E. COLI MUTATIONS Use the figure abo
ID: 151846 • Letter: G
Question
Gene Regulation in Prokaryotes – Lac Operon
E. COLI MUTATIONS
Use the figure above depicting a wild type operon (a) and examples of mutation (b-d) to answer the following questions.
(2 pts) You are studying lactose metabolism and have been culturing E. coli in different medias. One media has only glucose, one has only lactose, and a third has both glucose and lactose.
(1 pt) You notice that some of the cultures grown in the presence of both glucose and lactose are metabolizing lactose. Which type(s) of mutation(s) could cause this to occur?
(1 pt) You notice that some E. coli cultures failed to grow in the presence of only lactose. You know that normal E. coli are able to metabolize lactose so something must be wrong. Which mutation(s) could explain this?
(2 pts) Why is it advantageous for organisms to be able to regulate their gene expression? Please explain in 1-3 sentences.
(4 pts) You are studying how a certain species of grass decomposes in a semi-arid grassland in southern Arizona to get an idea on the C dynamics of the region. You find that after a few months the grass litter you’ve been sampling has a higher abundance of fungi than earlier samplings coupled with a decreased abundance of bacteria. You compare the chemical composition of the earlier litter samplings with the older litter and you find that this species of grass has a high concentration of cellulose and the cell solubles (sugars like glucose) have declined rapidly over the experiment.
(2 pts) You know that cellulase expression in microbial decomposers is contingent on Carbon Catabolite Repressor (CCR) which is a cellulose sensor and works the same as the lac repressor in E. coli. Would CCR be bound to the operator on the older litter? Please explain why in 1-3 sentences.
(2 pts) You find a healthy mixture of bacteria and fungi growing on the earlier litter samplings but you don’t find any bacteria on the older litter. What could explain these findings? Explain in terms of cellulose and cellulose monomers.
c) (repressor mutation) (a) I(wild type) Repressor binds cperator when the inducer is absent and forms an inducer- repressor complex when inducer is present Repressor protein mutation prevents repressor binding to the operator and produces constitutive synthesis of the lac operon lac repressor protein Allolactose Mutant protein (d) (super-repressor mutation) (b) O (operator constitutive mutation) Operator-site mutation prevents repressor protein binding and leads to constitutive synthesis of the lac operon Repressor protein mutation blocks binding to the inducer, preventing formation of the Mutant repressor protein binds to the operator, lac repressor proteinExplanation / Answer
1. I am studying lactose metabolism and have been culturing E.coli in different medias.
One media has only glucose:
All tests (a-d) will use go through glucose metabolism, because in this case lactose is absent. So, lac repressor binds tightly to the operator. Thus, RNA polymerase cannot attach to the promoter, and transcription will be prevented. So, there will be no lactose metabolism.
One media has only lactose:
Wild type (a) will go through the lactose metabolism. Because in this case, as lactose is present, allolactose binds to the lac repressor. So, repressor will make the operator free as well as the promoter, so transcription starts.
In case of (b), as operator site is mutated, it prevents repressor protein binding, so promoter will be free, will transcribe and allows lactose metabolism.
In case of (c), as repressor protein is mutated, it will not bind the operator, so promoter will be free, will transcribe and allows lactose metabolism.
In case of (d), mutant repressor binds to the operator, and it blocks binding to the inducer too, and promoter will not be free. So, transcription will not start, and lactose metabolism will not occur.
One media has glucose and lactose:
In this case, diauxic growth (double growth) will occur. This is a growth, occurs when two sugars are present in the media. In that case, the preferred sugar will be consumed first, in this case, glucose. Then a lag phase will come which will be followed by the other sugar metabolism, so as lactose.
In case of (a), wild type bacteria will metabolize all the glucose, and will start a speedy growth, then it will consume all the glucose, and will start metabolizing lactose.
In case of (b), lactose will be metabolized first, as operator site mutation prevents repressor protein binding and thus will lead to the constitutive synthesis of lac operon.
In case of (c), the same thing will happen, as in this case also, constitutive synthesis of lac operon will happen.
In case of (d), as lactose metabolism is not possible, only glucose metabolism will happen.
2. Some of the cultures grown in the presence of both glucose and lactose are metabolizing lactose. In this case (b) and (c) will be responsible. Because, in both the cases, constitutive synthesis of lac operon will happen.
3. Some E. coli cultures failed to grow in the presence of only lactose. So, we can say that case (d) will be fitted for this. As, repressor protein mutation blocks binding to the inducer and prevents formation of inducer-repressor complex. Mutant repressor protein binds to the operator here, and prevents transcription. So, bacteria will not be able to metabolize lactose. Thus they will not grow, as glucose is not present in the media, only lactose is present.
4. Cells must balance the benefit and cost of protein expression to optimize the fitness of the organism.The lac operon of the bacterium Escherichia coli is a model to qualify the physiological impact of costly protein production and to elucidate the resulting regulatory mechanisms. So, this is advantageous for organisms to regulate their gene expression.
Please ask multiple questions separately.
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