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Semmy [17]
4 years ago
14

Elijah wanted to learn more about the growth pattern of bacteria, so he performed the following experiment.

Biology
1 answer:
emmasim [6.3K]4 years ago
6 0

Answer:

attatch the images of the Petri dishes so we can further help in understanding this question

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The cell cycle has a regular system of checks and balances that prevents damaged or mutated cells from proceeding to the next ph
makkiz [27]

Answer:

A. True

Explanation:

Apoptosis is a normal cellular process in which the cell inherently programs the death of a mutated or damaged cell without affecting the healthy condition of the other surrounding cells by proliferating.

Apoptosis is a system of checks and balances that the cell uses to prevent damaged or mutated cells from proceeding to the next phase in the cell cycle. This curbs or prevent the damaged cell from passing on the problem to its daughter cells. Hence, according to this question, it is TRUE.

6 0
3 years ago
What body system functions in secreting chemicals that act as relatively slow-acting messengers between body parts? select one:
Sloan [31]
B The Endocrine system would be the answer
4 0
3 years ago
Determine whether each statement is true or false according to Darwin’s theory
sammy [17]

Yes Darwins Theory of the Survival of the fittest and the theory of Natural Selection are the most rightest statements.

Hope this HELPS !!!

5 0
3 years ago
Read 2 more answers
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

7 0
3 years ago
In the carbon cycle, all animals must consume
scoray [572]

Answer:

B is the anwser, I <em>think. </em>

Explanation:

1.  animals cannot perform photosynthesis

2. animals cannot go into competion with plants because plants are in a total different kingdom

3. Carbon is not in water,  yes there might be a charcol-carbon water filter, yet  carbon isn't in water

4 0
3 years ago
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