Using radioisotope to determine the blood volume A 0.20 mL sample a saline solut

Question

Using radioisotope to determine the blood volume A 0.20 mL sample a saline solution containing^59Fe is injected into an animal. Before the injection the sample has an activity of 1.0 times 10^4 cps. After a sufficient time for the blood to be thoroughly mixed, 0.20 mL of the blood is removed and its activity measured. If the blood sample shows an activity of 95 cps, calculate the blood volume of the animal. A blood-volume determination was carried out on a patient by injection with 20.0 mL of blood that had been radioactively labeled with Cr-51 to an activity of 4.10 mu Ci/mL. After a brief Period to allow for mixing in the body, blood was drawn from the patient for analysis. Unfortunately, a mixup in the laboratory prevented an immediate analysis, and it was not until 17.0 days later that a scintillation measurement on the blood was made. The radiation level was then determined to be 0.00935 mu Ci/mL. If^51Cr has t_1/2 = 17.7 days, what is the volume of blood in the patient?

Explanation / Answer

Solution:

(3) Given: 0.2 mL original sample containing 59Fe has an activity = 1 x 104 cps

And 0.2 mL recovered blood containing 59Fe sample has an activity = 95 cps

Therefore the ratio of the original sample and the recovered sample can give a measure of the change in activity and that can be correlated with volume of the blood in animal as follows.

original sample activity/ recovered blood sample activity = 1 x 104 cps/95 cps = 105.26

That means original sample (0.2 mL) is diluted to 105.26 times after injecting into the animal’s body.

Therefore the total volume of the blood in the animal’s body = 0.2x105.26 mL = 21 mL

(4) Given t1/2 = 17.7 days,

[A] = concentration of the isotope after 17 days of the experiment = 0.00935 Ci/mL

[A]o = concentration of the isotope immediately after drawing from the animal (Ci/mL)

We know radioactive materials follows 1st order kinetics, therefore we can write:

t1/2 = 0.693/k = 17.7 days (where, k = rate constant of radioactive decay)

or, k = 0.4 day-1

Now, we know for 1st order kinetics: kt = 2.302 log [A]o/[A]

Or, 0.4x17/2.302 = log [A]o/[A]

Or, log [A]o/[A] = 2.95

Or, [A]o/[A] = 102.95

Or, [A]o = [A]x 102.95 = 0.00935 x 102.95 = 8.33 Ci/mL

Now, we can say that:

4.10 Ci/mL radioactivity of the blood sample comprises = 20 mL blood

8.33 Ci/mL ------------------------------------------------------    = 20x8.33/4.1 mL blood

                                                                                                = 40.63 mL blood.

Therefore the patient has 40.63 mL of blood.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.