33-The GasPak anaerobic jar provides what environment? A. B. C. D. Oxygen free B
ID: 132913 • Letter: 3
Question
33-The GasPak anaerobic jar provides what environment? A. B. C. D. Oxygen free Bacteria free Carbon dioxide free All of the above Use the diagram below to answer the question 34 and 33 34- The letters on the Mueller-Hinton II agar plate indicate specific antibiotics that were tested using the Kirby- Bauer method. Which antibiotic is the most effective against the bacteria? a) G b) A c) E d) c 35-The clear area around the antibiotic disk where no bacteria growth is observed is referred to as a. b. c. d. Zone of inhibition Lawn of bacteria Zone of resistance Zone of sensitivity 36- After doing a urine CFU analysis, in which you transferred.01 ml of your urine sample onto an agar plate, you counted 3,700 colonies in your plate. The number of organisms per milliliter of sample is a) 370 cells/mL b) 370,000 cells/ml. c) 37000 cells/mL d) 37 cells/mlExplanation / Answer
Hi Answer:
Q. 33 The GasPak anaerobic jar provides what environment?
Answer: Option A [Oxygen free] In GasPak anaerobic jar provide oxygen free environment which is used for the growth of strict aerobic bacteria. Carbon dioxide is present in gaspak jars.
Q. 34. The letters on the Mueller-Hinton II agar plate indicate specific antibiotics that were tested using the Kirby-Bauer method. Which antibiotic is the most effective against the bacteria?
Answer: Option C [E] The antibiotic E shows the maximum zone of inhibition, it means it inhibit the growth of bacteria maximum.
Q.35. The clear area around the antibiotic disk where no bacteria growth is observed is referred as.
Answer: Option A [Zone of inhibition] The zone of inhibition means the effectiveness of the antibiotic which didn’t allow any bacterial growth. As larger the Zone of inhibition higher the effectiveness of antibiotic.
Q.36. After doing a urine CFU analysis, in which you transferred 0.01 ml of your urine ample onto an agar plate, you counted 3,700 colonies in your plate. The number of organisms per milliliter of sample is
Answer: Option B [370,000 cells/ml]
Because 0.01 mL means 10 microlitre, so in 10 microlitre no of colonies = 3700
Hence in 1 microlitre the number of colonies will be = 3700/10 = 370
Now 1 milliliter = 1000 microlitre
So. Number of colonies in 1000 microlitre= 370X1000= 370,000
Hence answer will be 370,000 cell/mL
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