A grad student has a culture of E. coli with a total volume of 1,800 mL, and OD reading at 550 nm of 1.7. He transfers 1 mL culture into 9 mL tryptone broth. Next, he transfers 0.1 mL of this dilution into 0.9 mL tryptone broth, and repeats this two more times to make a total of 4 serial dilutions. He then pipets 200 uL from the fourth dilution on an agar plate, spreads it over the surface, and puts the plate in an incubator to grow overnight. The next day, he counts 386 cfus (colony forming units) on the plate.

a. What is the dilution in the third tube? Answer with the exponent to the power of 10, that is ? in 10?
b. What is the dilution of cells spread onto the agar plate? Answer in decimal notation.
c. How many viable cells are present in the 1,800 mL starting culture? Answer in decimal notation.

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

1. a host loses nutrients to a parasite

2. a parasite only benefits when the host is alive (cause if the host dies the parasite dies with it )

3. producers to consumers

4. the tapeworm is a parasite

5. they convert energy from the sun (photosynthesis)

6. sunflower-robin (i think im 80% sure)

7. producer

8. consumer

9. prey

10. fresh water snails ( cause the green algea is the producer and the snails are the primary consumers and the small fish consumer so is the big fish)

substitution is the answer in this case