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NNADVOKAT [17]
3 years ago
11

A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of

laboratory rats a daily diet of precisely 125.6 units of niacin and 5,400 units of retinol. She uses two types of commercial pellet food. Food A contains 0.86 units of niacin and 30 units of retinol per gram. Food B contains 0.66 units of niacin and 40 units of retinol per gram. How many grams of each food does she feed this group of rats each day
Biology
1 answer:
julsineya [31]3 years ago
4 0

Answer:

- 100 grams of food A (x)

- 60 grams of food B (y)

Explanation:

According to this question, the group of laboratory rats are being fed a daily diet of:

- 125.6 units of niacin and 5,400 units of retinol

Let x represent the number of grams needed for food A.

Let y represent the number of grams needed for food B.

This means that:

Food A: Niacin = 0.86x, Retinol = 30x

Food B: Niacin = 0.66y, Retinol = 40y

Since there are 125.6 units of niacin fed to the rats in total, then;

0.86x (A) + 0.66y (B) = 125.6

Since there are 5400 units of retinol fed to the rats in total, then;

30x (A) + 40y (B) = 5400

Solving these two equations simultaneously using elimination method:

0.86x (A) + 0.66y (B) = 125.6 ........(eqn 1)

30x (A) + 40y (B) = 5400 ............ (eqn 2)

Multiply eqn 1 by 30 and eqn 2 by 0.86

30 × 0.86x + 0.66y = 125.6

0.86 × 30x + 40y = 5400

25.8x + 19.8y = 3768 ............ (eqn 3)

25.8x + 34.4y = 4644 ........... (eqn 4)

Subtract eqn 3 from eqn 4

14.6y = 876

Divide both sides by 14.6

y = 876/14.6

y = 60

Substitute value for y (60) into eqn 2

30x + 40y = 5400

30x + 40(60) = 5400

30x + 2400 = 5400

30x = 5400 - 2400

30x = 3000

x = 3000/30

x = 100

Therefore, she feeds the group of rats 100grams of food A (x) and 60 grams of food B (y).

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Complete question:

A population of wolves predates a population of moose on Isle Royale, Michigan, where there are fewer wolves than moose to start. The wolves prey on the moose and eat well, allowing them to have abundant offspring. However, as the wolf population rises, the moose population drops, and over time, the wolf population begins to drop also because of the reduced availability of resources. As the wolf population drops, moose are able to better survive and reproduce, causing the moose population to rise. With this abundance of moose, the wolf population is able to rebound until their population exceeds the moose population’s ability to support the number of wolves. Which population dynamic does this series of oscillations represent?

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Answer:

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Explanation:

The Delayed density dependence occurs when two species are in interaction (prey-predator or parasite-host), and the dynamic of each population follows the dynamic of the other population. The natality and mortality rates of one species follow the rates of the other species.

<em>Delayed density dependence dynamics regulate populations and allow the occurrence of equilibrium or balance in nature.</em>

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The prey population also affects the predator population. After the increase in the predator population size, the number of available prey decreases. The predator lives in an ideal environment but depends on the prey density. The more predators there are, the fewer prey there will be left. The predator population decreases exponentially due to the item's lack. The predation rate depends on density as well as natality and mortality rates.

So, to sum up,

  1. prey population increases in size
  2. predator population increases in size
  3. prey population decreases in size
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In this particular example, the wolves population follows the moose population and vice-versa. It is not simultaneous. It happens with a delay.

<em>Moose influences the wolf population because they are their source of food. The more moose, the more wolves there are. The fewer moose, the fewer wolves there are. </em>

This is a density-dependent dynamic because both populations are affected when they reach a high value.

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