Discussion: 1. Briefly explain the role each of these solutions has in the Gram

Question

Discussion: 1. Briefly explain the role each of these solutions has in the Gram stain procedure: a. Crystal Violet- b. Gram's lodine- c. Decolorizer- d. Safranin- 2. Which step of the Gram stain do you think is the most critical in differentiating between Gram positive and negative cells and why? 3. Nucleic acids are released when KOH is added to either Gram-positive or Gram-negative bacteria causing the resulting solution to appear stringy. Based on your knowledge of bacterial membrane structure, which of the two do you predict to give this result?

Explanation / Answer

1. a. Crystal violet is the primary stain which gives violet or purple color to the gram positive bacteria due to the thicker peptidoglycan layer, techoic acids and phosphate in the gram positive bacterial cell wall

b. Iodine is the Mordant which combines with the crystal violet and sets the dye in the cell

c. Decolorizer like acetone or ethanol is used to wash out the excess dye as well as primary stain from the thinner peptidoglycan cell wall of gram negative bacteria

d. Safranin is the counterstain or secondary stain which gives pink color to the gram negative bacteria due to less peptidoglycan content in the cell wall.

2. The most critical step is the Decolorizing step because if not properly done it may give erraneous results. Proper timing of decolorization stage is essential as if the step is done for a long time gram positive bacteria may appear as gram negative due to washing of the primary stain. When decolorizer is not given enough time to interact with the cell wall, gram negative bacteria still retain the purple color and appear as gram positive.

3. KOH is used as an alternative to staining technique when the staining results show ambiguity. KOH is used as an emusifying agent to react with a drop of the bacterial culture for about one minute.

Gram negative bacteria release the chromosomal material and results in the appearance of stringy precipitate of DNA. This is due to the thin peptidoglycan layer in gram negative cells and lack of techoic acids, phosphates which make them liable to the attack by KOH resulting in cell wall disruption and precipitate formation.

On the other hand, gram positive bacteria remains unchanged as the thick peptidoglycan layer prevents the disruption of cells by KOH

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