What is bacteria #2

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Answer 1

Answer:

Answer:

Bacteria are unicellular microorganisms that are not visible by the naked eye. they are found everywhere and exist in millions in population. some of them are harmful while some are essential for us like lactobacillus which helps in formation of curd.

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A scientist found that a Petri dish from an experiment had not been cleaned several days earlier. In the dish, she discovered a bacterial culture that she had not seen during the experiment. How could she gather evidence to support a claim about whether the bacteria had developed from the material left from the earlier experiment? a)Take a sample of the bacteria and grow it in an incubator. b)Gather dirty Petri dishes from other labs to see if they grow the same bacteria. c)Check all the clean Petri dishes for residue from the previous experiment. D)Try the original experiment again, do not clean some of the Petri dishes, and see if the bacteria grow.

What role does coral play in the ecosystem?

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Answer: Providing shelter and spawning grounds to a wide range of ocean life, coral reefs serve an important role in the marine ecosystem. ... Another role is protection from strong ocean currents and high waves. As the name "barrier reef" implies, reefs act as a barrier protecting the shorelines.

Hope this helps !

The force that pulls falling objects toward earth is called

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Answer:

The answer is gravity: an invisible force that pulls objects toward each other. Earth's gravity is what keeps you on the ground and what makes things fall. An animation of gravity at work. Albert Einstein described gravity as a curve in space that wraps around an object—such as a star or a planet.

A scientist is studying the lysogenic cycle of the lambda 1 phage of Escherichia coli. This means that she is investigating A. the integration and stabilizing of the lambda 1 phase into a host cell's genome.
B. the integration of the bacteria's genome with the viral genome outside of the bacteria's cell wall.
C. the integration of the viral genome into the ribosomes present in the bacteria to direct protein synthesis for the capsid formation.
D. the integration and stabilizing of a virus into its capsid, which provides protection until conditions are better for reproduction.

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Answer:

The correct answer is A. the integration and stabilizing of the lambda 1 phase into a host cell's genome.

Explanation:

Viruses have two reproductive cycle i.e, lytic and lysogenic cycle. During the lysogenic cycle, the viral genetic material gets incorporated into the bacterial DNA and gets replicated with the bacterial genome using bacterial molecular machinery.

During the lytic cycle, the viral protein is expressed which forms the progeny phages which release from the cell through cell lysis. Then the new phages infect other cells. Therefore during the lysogenic cycle, integration and stabilization of the viral genome in the host cell's genome occurs. Therefore the right answer is A.

Which cell is a plant cell?

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Answer:

Explanation:

This is a description of a(n)

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This refers to a comet

We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day, an individual of bacteria 1 can each 3 units of food A, 2 units of food B, and 1 unit of food C. An individual of bacteria 2 can each 1 unit of food A, 4 units of food B, and 1 unit of food C. An individual of bacteria 3 can eat 2 units of food A and food B but does not eat food C. In one day, the bacteria eat a total of 58 units of food A, 70 units of food B, and 20 units of food C. How many of each bacteria are there

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Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For food source A; we have:

3P₁ + P₂ + 2P₃ = 58 units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70 units of food B ---- (2)

For food source C; we have:

P₁ + P₂ = 20 units of food C ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

P₁ - 3P₂ + 0 = - 12

P₁ = -12 + 3P₂ ---- (4)

Replace, the value of P₁ in (4) into equation (3)

P₁ + P₂ = 20

-12 + 3P₂ + P₂ = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂ = 20

P₁ + 8 = 20

P₁ = 20 - 8

P₁ = 12

To find P₃; replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ = 7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

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