Not so simple math problem with inbreeding.

If we have as many males as females, how many parents of each gender are needed to limit inbreeding to 0.25% per generation?

Using the formula:
Ne = 4NfNm/(Nf+Nm)

And then:
DF=1/(2Ne)

So 25 males and 25 females gives us= Ne = 50 and DF= 1%

Dont know why but when i try and type in how i tried to calculate it it just crashes...

The very first step (in any problem really, not just this one) is to define your variables. Everything you've written is absolutely meaningless because we have no way of knowing what the variables stand for. In this specific case, I googled the formula and I am now prepared to make a reasonable assumption. If \(\displaystyle n_e\) is the effective population size, then your first formula is:

\(\displaystyle n_e=\dfrac{4 \cdot n_{\text{f}} \cdot n_m}{n_{\text{f}}+n_m}\), where \(\displaystyle n_{\text{f}}\) is the number of reproductive females, and \(\displaystyle n_m\) is the number of reproductive males.

However, I haven't even the slightest clue what "DF" might stand for in this context.

I have checked this http://www.fao.org/docrep/006/x3840e/X3840E04.htm very many times but i can still not figure out how to "find" the inbreed down to 0.25% per generation

problem is not finding it becouse there is a diagram showing it. Its just that i need to show how i calculated it.

Oh, okay. Well, given that new information, it seems as though you've worked through this problem on your own and arrived at the correct answer. Is there perhaps something that makes you doubt your answer?

Teacher told us that 100 female/ male is wrong.. so..


where: N_eF is the effective breeding number in an inbred population, and F is the average inbreeding in the population.

For some reason you've changed your last reply to this topic. For posterity's sake, the original is archived below:

Madmoremax said:

DF stands for the inbreed % in the population.

For example in 1/2x200 = 0.0025 = 0.25% = DF

100 females 100 males.

Click to expand...

One niggling detail before I continue. What you wrote is actually the following:


\(\displaystyle \dfrac{1}{2} \cdot 200\), but what I'm sure you meant was \(\displaystyle \dfrac{1}{2 \cdot 200}\). This would be typeset as 1/(2*200)

This appears to generate the correct answer. I'm not sure why your teacher says it's wrong.

Maybe he wants me to write down how i got to the right answer.





So what you are telling me that this is the right answer?
even if the teacher told me its the wrong answer? So confused right now.

ksdhart2 said:

For some reason you've changed your last reply to this topic. For posterity's sake, the original is archived below:




One niggling detail before I continue. What you wrote is actually the following:


\(\displaystyle \dfrac{1}{2} \cdot 200\), but what I'm sure you meant was \(\displaystyle \dfrac{1}{2 \cdot 200}\). This would be typeset as 1/(2*200)

This appears to generate the correct answer. I'm not sure why your teacher says it's wrong.

Click to expand...

No, hold on. Give me a second to figure some stuff out. I'm really having a hard time keeping up with this topic, due to all of the rapid-fire edits and new posts coming through. Using the very first formula you posted, the answer of 100 males and females is correct. But now you've changed the formula. I've skimmed some of the page you linked, and the original formula seems to be determining the amount of inbreeding in the next generation, whereas the new formula determines the long-term "settling down" of the inbreeding, accounting for the fact that an inbred generation will produce a next generation that's more inbred. Okay, so given that, let's work with the new formula. The effective breeding number in an inbred population is:

\(\displaystyle N_{\text{eF}}=\dfrac{N_{\text{e}}}{1+F}\)

where F is the average inbreeding in the population and is defined as:

\(\displaystyle F=\dfrac{1}{2 \cdot N_{\text{e}}}\)

where \(\displaystyle N_e\) is the effective breeding number and is defined as:

\(\displaystyle N_{\text{e}}=\dfrac{4\cdot N_{\text{f}}\cdot N_{m}}{N_{\text{f}}+N_{\text{m}}}\)

where \(\displaystyle N_{\text{f}}\) is the number of reproductive females and \(\displaystyle N_{\text{m}}\) is the number of reproductive males.

So, since we have a series of nested equations, the best approach is probably to start at the bottom and work your way up. You know from the problem statement that the number of reproductive males and reproductive females is equal. How does that information change the equation for \(\displaystyle N_{\text{e}}\)? Given that new value, what does the equation for F evaluate to? Given that, what does the equation for \(\displaystyle N_{\text{eF}}\) evaluate to? When will that value be 0.25%?

I wish i could say i understand Anything. My brain in blowing up at the moment. I´m having trouble with just the basic calculations.
Never bin so ashamed 36 years old and can´t even calculate this..

Madmoremax said:

I wish i could say i understand Anything. My brain in blowing up at the moment. I´m having trouble with just the basic calculations.
Never bin so ashamed 36 years old and can´t even calculate this..

Click to expand...

So lets say i have 200 Fishes. 100 Male and 100 Females.
Then the Ne = 200
F= 0,0025

1+F = 1.0025, Ne is 200 so, 200 divided with 1.0025 =0,0050125?

Madmoremax said:

So lets say i have 200 Fishes. 100 Male and 100 Females.
Then the Ne = 200
F= 0,0025

1+F = 1.0025, Ne is 200 so, 200 divided with 1.0025 =0,0050125?

Click to expand...

Not quite, but it's close. Your math is correct, but you're solving the wrong equation. What you've solved is 1.0025 divided by 200. Recall that in the formula, N_e (=200) is in the numerator of the fraction. What does that suggest you might do to fix the error? If you need a refresher on fractions, you might try here.

Thank you for trying to help.

After having 10 calculators up at the same time to try and figure this out. Its late i´m tired and frankly 5 hours trying to figure this Sh*t out.
i´m giving up.

Cheers