The image given below represents the biomagnification of DDT in an ... - Free Printable

Problem Analysis:

The image depicts a food chain in an aquatic ecosystem, showing the transfer of a toxic substance (DDT) through different trophic levels. The task is to determine the DDT concentration at each level and identify the values for A and B.

#### Key Concepts:
1. Bioaccumulation: Toxic substances like DDT accumulate in organisms as they move up the food chain. This means that higher-level consumers (e.g., predators) have higher concentrations of the toxin than lower-level organisms.
2. Concentration Factors: As you move up the food chain, the concentration of DDT increases because organisms consume multiple prey items, each containing some amount of DDT.

#### Given Data:
- Zooplankton: 0.04 ppm DDT
- Small Fish: 0.5 ppm DDT
- Fish-eating bird: 25 ppm DDT

We need to determine:
- A: The DDT concentration in the large fish.
- B: The DDT concentration in the water.

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Step-by-Step Solution:

#### 1. Understanding the Food Chain:
The food chain in the image is as follows:
- Zooplankton → Small Fish → Large Fish → Fish-eating bird

#### 2. Bioaccumulation Pattern:
From the given data:
- Zooplankton (lowest trophic level): 0.04 ppm DDT
- Small Fish: 0.5 ppm DDT
- Fish-eating bird: 25 ppm DDT

The increase in DDT concentration suggests a pattern of bioaccumulation. Let's calculate the concentration factor between each trophic level.

##### From Zooplankton to Small Fish:
The concentration of DDT in small fish is 0.5 ppm, while in zooplankton it is 0.04 ppm. The concentration factor is:
\[
\text{Concentration Factor} = \frac{\text{DDT in Small Fish}}{\text{DDT in Zooplankton}} = \frac{0.5}{0.04} = 12.5
\]

This means that small fish accumulate DDT 12.5 times more than zooplankton.

##### From Small Fish to Large Fish (Finding A):
Let the DDT concentration in large fish be \( A \). Using the same concentration factor (12.5), we can write:
\[
A = \text{DDT in Small Fish} \times \text{Concentration Factor} = 0.5 \times 12.5 = 6.25 \, \text{ppm}
\]

Thus, the DDT concentration in large fish (A) is 6.25 ppm.

##### From Large Fish to Fish-eating Bird:
The concentration of DDT in the fish-eating bird is 25 ppm, and the concentration in large fish is 6.25 ppm. The concentration factor is:
\[
\text{Concentration Factor} = \frac{\text{DDT in Fish-eating bird}}{\text{DDT in Large Fish}} = \frac{25}{6.25} = 4
\]

This means that the fish-eating bird accumulates DDT 4 times more than large fish.

#### 3. DDT Concentration in Water (Finding B):
Zooplankton are primary producers in this food chain and directly absorb DDT from the water. Since zooplankton have a DDT concentration of 0.04 ppm, and they are the first organism in the chain, the DDT concentration in the water (B) must be less than or equal to 0.04 ppm. Typically, the concentration in water is much lower, but for simplicity, we assume it is close to the concentration in zooplankton.

Thus, the DDT concentration in the water (B) is 0.04 ppm.

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Final Answer:

\[
\boxed{A = 6.25 \, \text{ppm}, \, B = 0.04 \, \text{ppm}}
\]