Which bird is best adapted to survive in an environment dominated by seed-bearing plants?
1 answer:
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Answer:
Explanation:
We'll represent the dilution as the volume transferred/total volume in the tube after transfer. So for the first dilution, 1 ml of culture was added into 9ml of broth bringing the total volume to 10 ml i.e 1/10
For the second dilution, 0.1 mL was added to 0.9mL of broth bringing the total volume to 1mL i.e 0.1/1.0 (this is also the same as 1/10). In relation to the stock culture, this is represented as 1/100 (that is 1/10 (first dilution) * 1/10 (second dilution)
Giving that the same was done for dilution 3 and 4,
third dilution is represented as: 1/10 and the fourth as 1/10 as well
In relation to the stock, dilution 3 is 1/1000
dilution 4 is 1/10000
a. the dilution in the third tube is or 1.0*
b. the dilution spread onto the agar plate is the fourth dilution, that is, or
c. Since the grad student got 386CFU/0.2mL (200 uL = 0.2mL),
the cells in 1mL is: 386/0.2 = 1930 = 1.93 *
the cells in 1800mL is therefore 1930*1800 = 3,474,000 = 3.4*10