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A \(1.00-\mathrm{mL}\) sample of blood is withdrawn from an animal, and the red blood cells are labeled with phosphorus-32 ( \(t_{1 j}=14.28\) days). The activity of this sample is 50000 disintegrations \(\cdot \mathrm{min}^{-1}\). The sample is then reinjected into the animal. A few hours later, another \(1.00\) -mL sample is withdrawn, and its activity is \(10.0\) disintegrations \(\cdot \mathrm{min}^{-1} .\) Determine the volume of blood in the animal. Assume that phosphorus-32 is uniformly distributed throughout the blood and that the activity due to phosphorus- 32 remains constant during the experiment. By using similar methods, it has been found that the human body contains about \(75 \mathrm{~mL}\) of blood per kilogram of body weight.

Short Answer

Expert verified
The total blood volume is 5000 mL.

Step by step solution

01

Understand the Problem

We start by understanding that we have two activities measured in different blood samples from the same animal. The first sample has an initial activity, and the second sample shows decreased activity after reinjection and mixing throughout the animal's blood. We aim to calculate the full volume of blood in the animal based on these activities.
02

Use the Activity Ratio

The activity ratio between the second and first samples represents the fraction of the blood volume the first sample constitutes regarding the total blood volume. Given activity for the first sample is 50000 disintegrations/min, and for the second sample, it's 10.0 disintegrations/min, we express this ratio as \( \frac{10.0}{50000} \).
03

Set Up the Equation for Blood Volume

If we denote the total blood volume in the animal as \( V_{total} \), we use the ratio from the second step:\[ \frac{1.00 \, \text{mL}}{V_{total}} = \frac{10.0}{50000} \].
04

Solve for Total Blood Volume

Rearrange the equation \( \frac{1.00}{V_{total}} = \frac{10.0}{50000} \) to solve for \( V_{total} \):\[ V_{total} = 1.00 \, \text{mL} \times \frac{50000}{10.0} \].
05

Calculate the Blood Volume

Now, calculate the blood volume, \( V_{total} \):\[ V_{total} = 1.00 \, \text{mL} \times 5000 = 5000 \, \text{mL} \].
06

Conclude with the Total Blood Volume

The total blood volume in the animal is determined to be 5000 mL based on the activities given and the volume of the sample injected and withdrawn.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Tracer Method
The radioactive tracer method is a powerful technique used to study the physiological processes in living organisms. This method involves using a radioactive isotope, often called a tracer, which is introduced into the subject. The tracer's journey through a biological system can be tracked because it emits radiation that can be measured with specialized equipment.
The primary purpose of using this method in the problem at hand is to estimate the blood volume of an animal. By labeling red blood cells with phosphorus-32, researchers can track how this isotope disperses and measures its activity across the animal's circulatory system.
  • Allows for precise measurement of blood volume
  • Enables understanding of physiological functions
  • Provides detailed insights without invasive procedures
This method's key advantage is that it permits accurate data collection while minimizing harm or disruption to the organism's natural processes.
Phosphorus-32 Decay
Phosphorus-32 ( ^{32} ext{P}) is a common radioactive isotope used in biological experiments due to its suitable half-life and decay properties. It is a beta emitter with a half-life of approximately 14.28 days.
Radioactive decay is a process where an unstable atom loses energy by emitting radiation. For Phosphorus-32, this involves emitting high-energy beta particles, which we can measure to determine the activity (disintegrations per minute).
  • Helps track biological processes
  • Used to label specific molecules, like red blood cells
  • Activity measurement helps in calculating concentrations and volumes
In this exercise, the decay of phosphorus-32 through beta emission allows researchers to estimate the blood volume by examining its uniform distribution and consistent decay rate over the period of the experiment.
Biological Half-Life
Biological half-life refers to the time it takes for half of a substance to be eliminated physically from a system, such as the bloodstream, due to biological processes. It’s important not to confuse this with the radioactive half-life, which is the time it takes for half of the radioactive atoms to decay.
Phosphorus-32’s radioactive half-life is essential for understanding how long we can measure its activity effectively in blood tissue. Knowing this value helps ensure that the activity measurements taken during the experiment are accurate and timely.
  • Different from radioactive half-life
  • Important for calculating dosing in medicine
  • Ensures effective timing of experiments
In this exercise's context, the experiments are designed within a time window that ensures the activity decline is negligible, allowing for precision in the measurements and calculations.
Activity Ratio in Blood Samples
The activity ratio is a critical concept when using radioactive tracers to measure blood volume. It involves comparing the amount of detectable radiation from two different samples. In this scenario, the first blood sample's activity is initially very high, as it directly carries the radioactive tracer injected into the bloodstream.
After mixing and redistribution throughout the body, the second sample's activity decreases due to dilution across the entire blood volume. This change in activity can be used to calculate the total blood volume because the ratio of these activities reflects how the tagged blood cells have been distributed.
  • First sample's high activity represents high concentration
  • Second sample's low activity signifies distribution
  • Ratio provides direct measure for estimating total blood volume
Through the measurement of this ratio, calculations can conclude the overall blood volume based on the principle that the decrease in activity in the second sample relates to the volume into which the first sample has been diluted.

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Most popular questions from this chapter

Strontium-90 is a radioactive isotope that is produced in nuclear explosions. It decays by \(\beta\) -emission with a half-life of \(29.1\) years. Suppose that an infant ingests strontium-90 in mother's milk. Calculate the fraction of the ingested strontium-90 that remains in the body when the infant reaches 74 years of age, assuming no loss of strontium-90 except by radioactive decay.

The following initial-rate data were obtained for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{3}(g)\) as described by the equation $$ \mathrm{N}_{2} \mathrm{O}_{3}(g) \rightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g) $$ \begin{tabular}{ccc} \hline Run & Initial pressure of \(P_{\mathrm{N}_{2} \mathrm{O}_{3}} /\) Torr & Initial rate of formation of \(\mathrm{NO}_{2}(g) /\) Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 1 & \(0.91\) & \(5.5\) \\ 2 & \(1.4\) & \(8.4\) \\ 3 & \(2.1\) & 13 \\ \hline \end{tabular} Determine the rate law for the reaction, expressed in terms of \(P_{\mathrm{N}_{2} \mathrm{o}_{3}}\) rather than \(\left[\mathrm{N}_{2} \mathrm{O}_{3}\right] .\) Calculate the value of the rate constant for the reaction.

Calculate the time required for the concentration to decrease by \(10.0 \%\) of its initial value for a firstorder reaction with \(k=10.0 \mathrm{~s}^{-1}\).

The reaction described by $$ \mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ is a second-order reaction with a rate constant \(k=\) \(0.54 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1} .\) How long will it take for \(\left[\mathrm{NO}_{2}\right]\) to be \(10.0 \%\) of its initial value of \(2.00 \mathrm{M} ?\)

Phosphine, \(\mathrm{PH}_{3}(g)\) decomposes according to the equation $$ 4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The kinetics of the decomposition of phosphine at. \(950 \mathrm{~K}\) was followed by measuring the total pressure in the system as a function of time. The following data were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction: \begin{tabular}{cc} \hline\(t / \min\) & \(P_{\text {total }} /\) Torr \\ \hline 0 & 100 \\ 40 & 150 \\ 80 & 167 \\ 100 & 172 \\ \hline \end{tabular} Determine the reaction rate law and calculate the value of the rate constant.

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