Gene control — practice questions

Comparison: Both prokaryotic and eukaryotic gene control involve turning genes 'on' or 'off' to adapt to cellular needs and environmental conditions. Both use proteins (repressors/activators in prokaryotes; transcription factors in eukaryotes) that bind to specific DNA sequences to regulate transcription. Both can lead to either increased or decreased production of specific proteins.

Contrast:

1. Organisation: In prokaryotes (e.g., E. coli), genes with related functions are often grouped into operons, allowing coordinated regulation. Eukaryotic genes are typically regulated individually, though genes in a pathway can be co-regulated by common transcription factors.
2. Chromatin Structure: Prokaryotic DNA is not associated with histones; eukaryotic DNA is packaged into chromatin, providing an additional layer of control. Eukaryotic gene expression is heavily influenced by epigenetic modifications like histone acetylation/methylation and DNA methylation, which alter chromatin accessibility.
   • Example: High histone acetylation in eukaryotes loosens chromatin, increasing gene expression, a mechanism absent in prokaryotes.
3. Regulatory Sequences: Prokaryotes use promoter and operator regions. Eukaryotes have more complex regulatory sequences, including distant enhancers and silencers, which specific transcription factors bind to.
4. Transcriptional Control: In the lac operon, both negative (repressor binding) and positive (cAMP-CAP binding) control at the promoter dictate transcription. In eukaryotes, transcription factors (activators/repressors) play a diverse role, forming complexes to recruit RNA polymerase II and mediate enhancer-promoter interactions.
   • Example: In low glucose, high cAMP activates CAP to boost lac operon transcription in E. coli. In eukaryotes, specific transcription factors would respond to a growth factor signal to activate transcription of genes for cell division.
5. Post-Transcriptional Control: Prokaryotes primarily rely on transcriptional control. Eukaryotes have extensive post-transcriptional mechanisms, including the crucial role of miRNAs in mRNA degradation or translational inhibition, which is absent in prokaryotes.
   • Example: miRNAs might downregulate the production of a specific protein in a human cell by binding to its mRNA, a regulatory step not seen in bacteria.
6. Nuclear Membrane: Eukaryotes have a nuclear membrane separating transcription and translation, allowing for more complex post-transcriptional processing and regulation. Prokaryotes lack this separation.

(a) Calculation of Fold-Induction:

• Step 1: Identify the basal level of expression (glucose only) and the fully induced level (lactose only).
  • Basal activity = 10 units
  • Induced activity = 4500 units
• Step 2: Calculate the fold-induction by dividing the induced level by the basal level.
  • Fold-induction = Induced activity / Basal activity
  • Fold-induction = 4500 units / 10 units = 450
• Answer: The lac operon is induced 450-fold.

(b) Explanation of Catabolite Repression:

The difference in activity is due to catabolite repression, where the presence of glucose prevents high-level expression of the lac operon.

• Step 1: Lactose-only medium (High Expression):

  • Lactose present: Lactose binds to the lac repressor, causing it to detach from the operator. This lifts the negative control.
  • Glucose absent: The absence of glucose leads to high intracellular levels of cyclic AMP (cAMP).
  • Activation: cAMP binds to the Catabolite Activator Protein (CAP). The cAMP-CAP complex then binds to the promoter region, acting as an activator. It greatly enhances the binding of RNA polymerase, leading to a high rate of transcription and high β-galactosidase activity (4500 units).

• Step 2: Glucose + Lactose medium (Low Expression):

  • Lactose present: As before, lactose binds to the repressor, lifting the negative control.
  • Glucose present: The presence of glucose inhibits the enzyme that produces cAMP, resulting in very low intracellular cAMP levels.
  • Lack of Activation: Without sufficient cAMP, the cAMP-CAP activator complex does not form. RNA polymerase can still bind to the promoter (since the repressor is gone), but its binding is weak without the help of the activator. This results in a very low rate of transcription and therefore low β-galactosidase activity (450 units).