Can anyone help me with this problem of toxicology?

I have to calculate lethal dose 50 (LD50):

number of animals in each group: 10

GROUP DOSE ANIMALS DEATH

GROUP1 20 mg/kg 0
GROUP 2 40 mg/kg 1
GROUP 3 80 mg/kg 2
GROUP 4 160 mg/kg 7
GROUP 5 360 mg/kg 10


thank you

Roughly around 120, there is a way to calculate it exactly but its been too long since i did this kind of maths for my degree.

yeah i have to calculate with "probits method", but thank you!, anyway if you know how to calculate it exactly this is my mail: hispano-suizo@hotmail.com.


thanks for posting

make a plot , find the dose at which 5 rats died and thats your LD50.

yeah i have to calculate with "probits method", but thank you!, anyway if you know how to calculate it exactly this is my mail: hispano-suizo@hotmail.com.


thanks for posting

Oh ok so you need to convert mortality to probabiltiy units first. I will try and help you later this evening.

Oh ok so you need to convert mortality to probabiltiy units first. I will try and help you later this evening.

thank you very much, yeah the mortality in each group it would be:

1)0%
2)10%
3)20%
4)70%
5) 100%

I have emailed you a spreadsheet. Let me know if you have any questions. The LD50 is 133 mg/Kg.

allez to toxicology

I have emailed you a spreadsheet. Let me know if you have any questions. The LD50 is 133 mg/Kg.

Thank you very much friend, I have received your mail:


I have understood all, the only question i have is: (sorry about my english)



when i calculated the probits (in statistic tables) i got:

0% = 0.5
10% = 3.72
20% = 4.16
70% = 5.52
100% = 99.5


your results of probits are very similar of mine (except 100% I have no idea why I got 99.5 it's not logical..). But I understund it better with your spreadsheet, so thank you very much!!!!

I calculated probits using the erf in MATLAB. Also the dosage corresponding to 0 death should never be used to calculate LD50 since technically there would be no death when dosage = 0 , 10 or even 30 mg/kg

I calculated probits using the erf in MATLAB. Also the dosage corresponding to 0 death should never be used to calculate LD50 since technically there would be no death when dosage = 0 , 10 or even 30 mg/kg

aham... yeah it's logical (0% shouldn't be used to calculate LD50...), the problem I only have now is that when I calculate it using formulas i got LD50 = 78.542 mg/kg (logically it's wrong), so I'm going to try no calculate using dosage = 0

See you soon..haha :wavey: thanks again

I calculate again...now my result of LD50 is 83.18 mg/kg haha


just_me, can I send you a mail with numeric counts and maybe can you look it? (sure I have some errors and I don't know) If you have no time, no probs!!

Thats the way you work it out, its by making a graph and then reading off the ED50 value.

Brings back memories.

I think i used to do EC50 no idea what it stood for.

I calculate again...now my result of LD50 is 83.18 mg/kg haha


just_me, can I send you a mail with numeric counts and maybe can you look it? (sure I have some errors and I don't know) If you have no time, no probs!!

Send me an email and i'll look at it.

I calculate again...now my result of LD50 is 83.18 mg/kg haha


just_me, can I send you a mail with numeric counts and maybe can you look it? (sure I have some errors and I don't know) If you have no time, no probs!!

The problem with your calculation is that you are fitting your data assuming the relationship between probits and log(dosage) to be linear


y= alog(x)+b

when y = 5 , x =LD50

LD50 = antilog((5-b)/a)

But the relationship as you can see from the plot is not linear. If you were to drop the pt corresponding to doage of 40 you will get a value just above 100. Why are you assuming a linear relationship ?

The problem with your calculation is that you are fitting your data assuming the relationship between probits and log(dosage) to be linear


y= alog(x)+b

when y = 5 , x =LD50

LD50 = antilog((5-b)/a)

But the relationship as you can see from the plot is not linear. If you were to drop the pt corresponding to doage of 40 you will get a value just above 100. Why are you assuming a linear relationship ?



Aham, it's true, it's not a linear relationship. Maybe in class the teacher was wrong..i dunno...I'm student of last year of veterinary and I just remember maths....thank you just_me for help, if you go to spain someday, that's sure I will invite you a drink hehe