A. Subject taking 2 liters of water per day and have body weight of 50 kg.
B. Subject taking 2 liters of water per day and have body weight of 70 kg.
C. Subject taking 1 liters of water per day and have body weight of 10 kg.
D. Subject taking 5 liters of water per day and have body weight of 100 kg.
E. Subject taking 1 liters of water per day and have body weight of 70 kg.
Choose the correct answer from the options given below:
Toxic Substance Daily Dose Calculation
This question asks us to compare the daily intake of a toxic substance across different subjects. We are given that the substance is present in water at a concentration of 10 mg/L, which is also known as 10 ppm. To compare the exposure levels accurately, we need to calculate the dose for each subject in terms of milligrams of substance per kilogram of body weight per day (mg/kg/day).
Daily Dose Calculation Explained
The daily dose is calculated by considering how much water a person drinks and their body weight. A higher dose means a greater exposure relative to body size.
The formula to calculate the daily dose is:
$$ \text{Daily Dose (mg/kg/day)} = \frac{\text{Concentration of substance in water (mg/L)} \times \text{Water intake (L/day)}}{\text{Body weight (kg)}} $$
In this problem, the concentration is constant at 10 mg/L for all subjects.
Subject Daily Dose Calculations
We will now calculate the daily dose for each subject (A, B, C, D, and E) using the formula. The concentration of the toxic substance in water is 10 mg/L.
| Subject | Water Intake (L/day) | Body Weight (kg) | Calculation | Daily Dose (mg/kg/day) |
|---|---|---|---|---|
| A | 2 | 50 | $$ \frac{10 \text{ mg/L} \times 2 \text{ L/day}}{50 \text{ kg}} $$ | $$ \frac{20}{50} = 0.4 $$ |
| B | 2 | 70 | $$ \frac{10 \text{ mg/L} \times 2 \text{ L/day}}{70 \text{ kg}} $$ | $$ \frac{20}{70} \approx 0.286 $$ |
| C | 1 | 10 | $$ \frac{10 \text{ mg/L} \times 1 \text{ L/day}}{10 \text{ kg}} $$ | $$ \frac{10}{10} = 1.0 $$ |
| D | 5 | 100 | $$ \frac{10 \text{ mg/L} \times 5 \text{ L/day}}{100 \text{ kg}} $$ | $$ \frac{50}{100} = 0.5 $$ |
| E | 1 | 70 | $$ \frac{10 \text{ mg/L} \times 1 \text{ L/day}}{70 \text{ kg}} $$ | $$ \frac{10}{70} \approx 0.143 $$ |
Decreasing Daily Dose Order
To arrange the subjects in decreasing order of their daily dose, we compare the calculated values:
- Subject C: 1.0 mg/kg/day
- Subject D: 0.5 mg/kg/day
- Subject A: 0.4 mg/kg/day
- Subject B: approx. 0.286 mg/kg/day
- Subject E: approx. 0.143 mg/kg/day
Based on these calculations, the subjects arranged in decreasing order of their daily dose are C, D, A, B, E.
Important Questions from Toxic Chemicals - Teaching
- Match the LIST-I with LIST-II
LIST-I LIST-II A. Dichlorvos I. Type A organophosphate insecticide B. Melathion II. Type B organophosphate insecticide C. Parathion III. Type C organophosphate insecticide D. Lindane IV. Organochlorine Insecticide
Choose the correct answer from the options given below: - Which of the following is most important analog of DDT that have same general size, shape and possess the same insecticidal properties but is more biodegradable and less bioaccumulative ?
- Given below are two statements. One labelled as Assertion (A) and the other labelled as Reason (R) :
Assertion (A) : The fundamental goal of a dose - response assessment is to obtain a mathematical relationship between the amount of a toxicant that a human is exposed to and the risk that there will be.
Reason (R) : To apply dose - response data obtained from animal bioassay to humans, a scaling factor must be introduced. Choose the correct answer : - Which of the following statement is incorrect ?
- The most common form of lead present in pesticide is :
