Yeast counterselection — 5-Fluoroorotic acid (5-FOA) inhibits growth if which selectable marker gene is present (functional) in yeast?
Correct Answer: URA3
Introduction / Context:5-Fluoroorotic acid (5-FOA) is a cornerstone reagent for yeast genetics. It enables efficient counterselection against URA3+ cells, facilitating marker recycling and precise genome engineering via homologous recombination.
Given Data / Assumptions:
- URA3 encodes orotidine-5′-phosphate decarboxylase in uracil biosynthesis.
- URA3 converts 5-FOA to a toxic product (5-fluorouracil), killing URA3+ cells.
- URA3- cells survive on 5-FOA medium when supplemented with uracil, enabling selection for URA3 loss.
Concept / Approach:Counterselectable markers rely on gene products that metabolize harmless analogs into toxins. URA3’s activity on 5-FOA is the canonical example used to pop out integrated cassettes or select against background URA3 activity after cassette shuffling.
Step-by-Step Solution:
Match 5-FOA toxicity to URA3 enzymatic conversion.Eliminate HIS3, LYS2, TRP1, LEU2 since they lack this conversion.Choose URA3 as the correct marker for 5-FOA counterselection.Verification / Alternative check:Standard gene deletion cassettes (e.g., URA3 pop-in/pop-out) leverage 5-FOA plates to isolate URA3- recombinants.
Why Other Options Are Wrong:
- a/b/d/e: These markers do not render cells sensitive to 5-FOA.
Common Pitfalls:Forgetting that URA3+ is lethal on 5-FOA unless uracil synthesis is bypassed; always supplement appropriately during selections.
Final Answer:URA3