Lac Operon Partial Diploids: How to Solve Merodiploid Problems

A partial diploid (merodiploid) is a bacterium with its normal chromosome plus a second copy of part of the lac operon on a plasmid, written with a slash, for example I⁺ O⁺ Z^- / I^- O⁺ Z⁺. These questions look intimidating, but they reduce to one question per gene: does the mutation act only on its own DNA (cis) or on both copies (trans)?

The parts and what each one does

• lacI encodes the repressor, a diffusible protein. It binds the operator and blocks transcription until allolactose binds it.
• lacO is the operator, a DNA site. It only controls the genes on the same DNA molecule.
• lacP is the promoter. Like the operator, it works only on its own DNA.
• lacZ encodes beta-galactosidase and lacY encodes the permease.
• CAP-cAMP is a positive activator. cAMP is high only when glucose is low, so the operon is strongly on only when lactose is present and glucose is absent.

Cis versus trans: the whole trick

A product that diffuses (repressor, enzyme) can work on either copy, so a good copy of that gene can rescue a bad one in trans. A DNA site (operator, promoter) cannot, so its defect stays on its own molecule.

A four-step method

1. Write each chromosome on its own row so you can see which genes share a molecule.
2. Ask whether a working repressor exists anywhere. If any I⁺ is present, repressor is made for both copies (unless I^s is also there).
3. For each copy, ask whether its own operator and promoter are intact. Only then can that copy's Z and Y be transcribed.
4. Add lactose and glucose conditions. Without lactose, only a copy with a mutant operator (or no working repressor) is on. With lactose, a copy is on if it has an intact promoter and, for high levels, glucose is low.

Worked examples

Example 1: I⁺ O⁺ Z⁺ / I^- O⁺ Z^-. One good repressor is made and acts on both operators. Without lactose, both operons are off; with lactose, the Z⁺ copy is on. Regulation is normal.

Example 2: I⁺ O^c Z^- / I⁺ O⁺ Z⁺. Repressor is plentiful, but the O^c copy ignores it. That copy is on constitutively, yet it carries Z^-, so it makes no working enzyme. The Z⁺ copy has a normal operator and stays repressed without lactose. Result: no beta-galactosidase without lactose, because the constitutive operator is cis to the broken gene.

Example 3: I⁺ O^c Z⁺ / I⁺ O⁺ Z^-. Now the constitutive operator sits cis to a good Z. Beta-galactosidase is made even with no lactose.

Example 4: I^s / I⁺. The super-repressor cannot bind allolactose and stays on the operator, so the operon cannot be induced even with lactose.

Low but detectable expression with glucose only

With glucose as the only carbon source and no lactose, a normal operon (I⁺ O⁺ P⁺) shows a low, detectable basal level of transcription because repression is not perfect. Genotypes that are constitutive (I^- or O^c with working Z) give higher mRNA, but glucose still keeps cAMP low, so even they are below their maximum. Pick the genotype with a working repressor and operator when the question describes low-but-detectable levels.

Common traps

• Treating the operator as diffusible. O^c is cis-acting only.
• Forgetting glucose. Without CAP-cAMP, expression stays low even with lactose.
• Reading I^s as recessive. It is dominant.
• Ignoring which copy carries the good Z. Constitutive does not help if the constitutive copy has the broken enzyme gene.

One-minute recap

• Diffusible products (repressor, enzymes) act in trans; DNA sites (operator, promoter) act in cis.
• O^c is constitutive only for genes on its own DNA; I^s is dominant and uninducible.
• Maximal expression needs lactose present and glucose low.