Question

Partial diploids can be produced in the bacterium \(\mathrm{E}\). coli for the lac operon. As a consequence, some mutations can be bypassed in terms of their disruption of the operon by production of the partial diploids with wild-type alleles at the sites of mutation. In each case the wild-type allele is dominant to its mutant homologue. Determine for each of the following partial diploids whether enzyme production will result constitutively or inducibly (normally) when the inducer molecule is introduced: (a) \(\underline{i}^{\pm} \mathrm{o}^{\pm} \underline{z}^{-}-\mathrm{y}^{-} \mathrm{a}=\) \(\mathrm{i}^{+} o^{c} \mathrm{z}^{+} \mathrm{y}^{+} \mathrm{a}^{+}\) (b) \(\underline{i}=\underline{o}^{\pm} \underline{z}^{\pm}-\mathrm{y}^{\pm} \underline{a}^{\pm}\) \(\mathrm{i}^{+} \mathrm{o}^{+} \mathrm{z}^{+} \mathrm{y}^{-} \mathrm{a}^{-}\) (c) \(\underline{i}^{\pm} \underline{o}^{\mathrm{c}} \underline{z^{\pm}}-\mathrm{y}=\underline{\mathrm{a}}=\) \(i^{-} o^{+} z^{-} y^{-} a^{-}\)

Short answer

For the given partial diploids, enzyme production will be: (a) Inducible (normal) (b) Constitutive (always active) (c) Constitutive (always active)

Step-by-step solution

(a) Given partial diploid sequence

The given partial diploid sequence is: \(i^{\pm} o^{\pm} z^{-}-y^{-}a=\) \(i^{+} o^{c} z^{+} y^{+} a^{+}\) Here '-' represents transcriptional fusion.

(a) Analyzing partial diploid sequence

In this case, the first part has a mutant operator \(o^{\pm}\) and thus does not allow enzyme production. However, the second part, with wild-type operator \(o^{c}\) and all other wild-type genes, will have normal (inducible) enzyme production.

(a) Determining enzyme production

For partial diploid (a), enzyme production will be inducible (normally) when the inducer molecule is introduced.

(b) Given partial diploid sequence

The given partial diploid sequence is: \(i=\underline{o}^{\pm} \underline{z}^{\pm}-\mathrm{y}^{\pm} \underline{a}^{\pm}\) \(\mathrm{i}^{+} \mathrm{o}^{+} \mathrm{z}^{+} \mathrm{y}^{-} \mathrm{a}^{-}\) Here '-' represents transcriptional fusion.

(b) Analyzing partial diploid sequence

In this case, enzyme production occurs constitutively (always active) due to the mutant operator \(o^{\pm}\) in the first part. The second part will not allow enzyme production since both \(y^{-}\) and \(a^{-}\) are mutant.

(b) Determining enzyme production

For partial diploid (b), enzyme production will be constitutive since there is an active mutant operator \(o^{\pm}\) even though some genes are in the wild-type form.

(c) Given partial diploid sequence

The given partial diploid sequence is: $i^{\pm} \underline{o}^{\mathrm{c}} \underline{z^{\pm}}-\mathrm{y}=\underline{\mathrm{a}}=$ \(i^{-} o^{+} z^{-} y^{-} a^{-}\) Here '=' represents plasmid fusion.

(c) Analyzing partial diploid sequence

In this case, enzyme production occurs constitutively (always active) due to the mutated regulator gene \(i^{-}\). The wild-type operator \(o^{+}\) is dominant and thus is present. The genes responsible for enzyme production (\(z^{\pm}\), \(y\), and \(a\)) are all mutated in both parts of the partial diploid and do not contribute to enzyme production.

(c) Determining enzyme production

For partial diploid (c), enzyme production will be constitutive due to the mutated regulator gene \(i^{-}\) in both parts of the sequence.

Key concepts

Lac Operon Regulation

The lac operon system is a brilliant example of genetic regulation in Escherichia coli, an extensively studied bacterial organism. It consists of three structural genes, lacZ, lacY, and lacA, which are involved in lactose metabolism. These genes are regulated collectively by a set of control sequences: the promoter (p), the operator (o), and the regulator (i).

How does this work? Normally, when lactose is not present, a protein product of the i gene acts as a repressor, binding to the operator and blocking transcription of the lac structural genes. As lactose enters the picture, it acts as an inducer by binding to the repressor and causing it to change shape and detach from the operator. This inducement allows RNA polymerase to transcribe the structural genes, thus enabling the cell to produce enzymes to metabolize lactose.

An interesting experiment involves creating partial diploids, which are cells carrying two copies of certain genes, one on the chromosome and another on an extra fragment of DNA. These diploids help scientists unravel complex regulation pathways, especially when one copy of the gene is mutated. By introducing wild-type alleles, researchers can observe the effects of certain mutations and determine whether they result in constitutive (constant) or inducible (controlled) enzyme production.

Constitutive Enzyme Production

Enzymes are the workhorses of the cell, facilitating biochemical reactions. Constitutive enzymes are synthesized continuously, regardless of the environmental conditions. This means their production is unregulated; they are produced at a constant rate without the need for an inducer.

In the context of the lac operon, constitutive enzyme production occurs due to mutations that affect the repressor binding site on the operator gene (o) or the repressor gene itself (i). These mutations can prevent the repressor from binding to the operator, which leads to the continuous transcription of the lac genes. This type of mutation is illustrated by a partial diploid that has a wild-type allele paired with a mutant operator - the regulatory system fails to respond to the absence of lactose, and the enzymes are produced all the time.

Inducible Enzyme Production

Opposite to the constitutive system, inducible enzymes are produced only in the presence of specific inducers. This adaptive system allows bacteria to save energy by only producing proteins when they are beneficial to the cell. In the lac operon, lactose acts as an inducer. It binds to the repressor protein and prevents it from repressing gene transcription, thus allowing the structural genes to be expressed.

Inducible enzyme systems, like the lac operon, are a prime example of an organism’s ability to adapt to environmental changes. When glucose is scarce and lactose is available, the inducer initiates the breakdown of lactose by activating the necessary genes. This regulation helps the cell optimize resource use.

Bacterial Genetics

Bacteria have sophisticated systems to ensure their survival, and scientists use these systems to uncover the principles of genetics. Bacterial genetics involve the study of heredity and variation in bacteria. Unlike eukaryotic organisms that have multiple chromosomes, bacteria usually have a single circular chromosome, but they can also carry small, circular DNA fragments called plasmids.

Techniques like creating partial diploids are useful in bacterial genetics for studying gene regulation and function. Bacterial genetics not only furthers our understanding of basic biological processes but is also invaluable in biotechnology, allowing us to harness bacteria for the production of pharmaceuticals, bioremediation, and other applications. The principles learned from bacteria such as the lac operon's regulation mechanisms have far-reaching implications in understanding and manipulating complex biological systems.