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half life
In my assay of 50uL of 125-I-insulin I have 20,000 cpm at the expiry date. The half life of 125-I-Insulin is 60 days. So how can I calculate how many counts a 50uL sample of 125-I-labelled insulin would contained at 4 months past the expiry date?
cheers |
125-l degradation is a First Rate Chemical reaction?
This is what I know so far:
Half life time for 125-l insulin = 60 days Original starting amount of 125-l insulin = 50 microliters cpm is counts per minute At expiry date = 20,000 counts per minute 4 months approximately 31 days, 30 days, or 29 days 4 months (30 days/ 1 month) = 120 days This is what we are seeking for: counts per minute at 4 months past the expiry day. That is beyond the 60 days half life. We might need to make a graph to figure out the count at 120 days past the expiry date. The expiry date is not the same as the half life. This problem is based upon the first order chemical reaction rate laws. I will need a few nights to do it. I do not have your text book to study the examples. You are making it harder on the answerer than yourself. You would be better to study the examples in your book carefully than to asks me. It might be a problem that the class wrote. I have thought about a Michaelis-Menten kinetics, but we are not studying the reaction substrate and an excess of enzyme. This is not exactly an initial reaction rate and concentration problem in a fixed amount of reactants and resulting products. The question portrayed itself as a half life, first order chemical reaction in the degradation of 125-l insulin. It is one reactant degrading into another product, thus lowering the counts per minute. Half life reaction was discovered by experimentally measuring the counts per minute. There are many computational chemistry situations where it has to be experimentally measured, before using descriptive laws and mathematics techniques that I learned in computational chemistry. It is often referred as chemistry & computers. It should be taken, before trying physical chemistry. It is viewed as difficult, but not true, just carefully read the problems and keep an open mind and try to solve it with some math and treat it like word problems and asked yourself does the answer relates? Do the units cancel out? It is relating to the real world occurances? I do not enjoy world problems, but it is as fun as puzzle solving. I enjoy the puzzle feeling that is why I am trying this. The dinner was great at my corner. I am still fascinated with photography, therefore I am sane. A Ph.D, that is too much financial liability, what if I cannot defend my thesis? What could I invent that is so original that the whole world would come and see? I will own my own soap, sodium, and halogen factory. Oh, no I am still in earth science. I cannot assure the correctness of this problem. Here are my findings. I suspect this 125-l insulin degrades in a first order reaction. I have not done any studies on the mechanism of this degradation of 125-l insulin. I am ignorant. As the chemicool.com says, we will solve it for you. That is vague. Who? Will it be right? It is scary, I cannot answer you. I did not write this in an equation editor, so you will have to follow as best as you can by text and ( ) grouping symbols for one term, denominator or numerator. * is the multiplication symbol, just like in computer programming. R=kN k=constant R=rate N=number, I will substitute this with counts per minute ln(Nt/No)=-kt t=time interval of decay k=decay constant Nt=remaining number or remaining counts per minute Change of concentration with time is part of this first order reaction Rate= - (delta [ A concentration]/delta time) = k[A concentration] ln[A] t - ln[A] = -k*time ln([A]t/[A]0) = - kt The Half Life ln (1/2[A]0/[A]0)=-k time LN 1/2 = -kt t=-(LN1/2)/k = 0.693/k Half Life = 60 days solve for K 60 days is the half life t=-(LN1/2)/k=0.693/k 60 days = 0.693/k 60 days * k = 0.693 k=0.693/60 = 0.0116 k=0.0116 50 micro liters is extra information, because it was not used in the calculation. Even after expiration time, we still have 50 micro liters, but less counts per minute, because the 125-l insulin in this 50 micro liters have degraded. I do not know this for sure, because I am not in your class or in your job or project. I am doing this blindly. If there are errors, it is understandable. All I have is a BS in earth science, so please. My speciality is the minerals that came out of the melt in volcanos and the sediment patterns and lithology of alluvial fans. Here is how we can check if 0.0116 is the constant? t half life = 0.693/0.0116 = 59.7 days = 60 days ln (Nt/N0) = -kt= -(0.0116 days^-1) (t) t=120 days after expiration Here is my troubled step. I know the counts per minute after expiration, but I do not know at initial or at half life. I am guessing from here on. t=120 days after expiration N0=20000 counts per minute What is Nt @ 120 days after expiration ln(Nt/N0) = -kt= -(0.0116 days ^-1) * (120 days) ln(Nt/N0) = -kt = -1.3920 Nt/N0=e^-1.3920=4.0229 N0=20000 counts per minute Nt=? Nt=4.0229 *N0=(4.0229)(20000 counts per minute) Nt=80457.7558 counts per minute Nt=80458 counts per minute This does not seem correct, because we should have less and not more. This higher count than 20000 counts per minute indicates it was before the expiration date We will have to reject this answer and keep the k=0.0116 days^-1 This is expected, because I am not in your class and I do not have your book or attended the lecture. I do not know who or where, typical. The quesions posted by high school, elementary school, and middle school are viewed more than this one. I am not a teacher. I am turtle earth scientists, telescope builder, camera man, microscope repair man, and lives near a corner. Let us try again: ln(Nt/N0) = -kt = -(0.0116 days^-1) (60 days)=-0.6960 (Nt/N0)=e^-0.6960=0.4986 N0=? Nt=20000 counts per minute 20000 counts per minute/N0=0.4986 20000 counts per minute = 0.4986 * N0 20000 counts per minute/0.4986 = N0 40000 counts per minute = N0 When the 125-l insulin was new, it was 40000 counts per minute Expiration date it was 20000 counts per minute as given from your message. 60 days +120 days, 4 months after the expiration date= 180 days ln(Nt/N0) = -kt =-(0.0116 days^-1) (180 days)=-2.0880 (Nt/N0) = e^-2.0880=0.1239 N0=20000 counts per minute Nt=? That is your answer to this question. What is the counts per minute 120 days or 4 months after the expiration date? (Nt/20000 counts per minute at the expiration date)=0.1239 Nt=01239 * 20000 counts per minute Nt=2478.6952 counts per minute Nt=2479 counts per minute This answer seems correct, but I am not sure. I am not responsible, if you failed school. I am a turtle. Leave me to crawl back into my refuge area. 2479 counts per minute is less than 20000 counts per minute. The time is beyond half life, expiration and initial. It might be right. This is a hard problem. I still do not know what to do with the 50 micro liter fact. You know what to do with that? |
125-l insulin problem might be a First OrderChemicalReaction
This problem might be a first order reaction. I am not sure how to use the 50 micro liters. It appears to be the same volume in starting, initial and ending, final. I think it is to affirm that there is not loss in the volume or mass, to confirm to the law of mass conservation, a closed system. The insulin might be in solution, but degradation within the solution medium. The solution is an inert medium to store it for delivery, use, and storage. I am not sure. Only you know about what is the 50 micro liters' purpose.
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