# How to predict future generations from heritability

The heritability of human intelligence is estimated to be around 0.5. There are of course various estimates, some lower, some higher. But let's work with this value of 0.5 for the moment.

I wanted to know how this relates to the trait value of future generations and read a bit about it. For example, in some related posts:
how should one interpret heritability is it related to r2
why is a heritability coefficient not an index of how genetic something is
how to interpret the breeders equation

Turns out the breeders equation gives us an idea about that. But we need a second variable: Selection. I read that selection is based on the reproductive success of individuals with a certain trait value. But how would I calculate this value? Let's assume that the reproductive success is higher, for humans with higher intelligence. Let's simplify and say, the fertiliy rate of above average indivdiduals is 2.10 and the same value for below average individuals is 1.90. How would I plug those numbers into the breeders equation?

I tried the other way around. We know that intelligence is increasing over generations in western civilizations (see Flynn effect). Again, estimates vary (and can actually be negative) but we can work with an increase of 2 IQ points per decade. So lets say 5 IQ points increase from one generation to the next in a western civilization population. That would mean:

S = R/h²
S = 5/0.5 = 10

What would this value of 10 mean? How does this boil down to the reproductive success of individuals with certain trait values?

## 1 Answer

"How would I plug those numbers into the breeders equation?" You would use those fertility rates in calculating the selection differential (S), which would essentially be a weighted intelligence average of the reproducers minus the population average (reproducers+non-reproducers). So, if the population starts wih 100 IQ and those with 125 IQ produce 2.1 and those with 75 IQ produce 1.9 and those two groups are equally populated, the weighted intelligence avg of reproducers becomes (125*2.1+75*1.9)/4=101.25. The selection differential is then 101.25-100=1.25. Response is 1.25*.5=.625. IQ moves to 100.625. In your example, S=10. That is, (x*2.1+y*1.9)/4-100=10, where x is the avg IQ of the above average group and y is the avg IQ of the below average group, assuming the two groups are equally populated.

• Thanks. I got a slightly different result. $$S=\frac{1}{\bar w}\sum_{i=1}^{\text{nb inds}}m_iw_i - \bar m$$ $$m_i = 125 \quad w_i = 2.1 \quad | i=1$$ $$m_i = 75 \quad w_i = 1.9 \quad | i = 2$$ $${\bar w} = 2$$ $${\bar m} = 100$$ $$S=\frac{1}{2}((125*2.1-100)+(75*1.9-100)) = 102.5$$ – vanao veneri Nov 21 '16 at 9:51
• I just realized that this might be due to the calculation taking $w_i$ as the value for couples. So for individuals we would divide by two. But then we would also have to divide ${\bar w}$ by two, and we get the same result. Maybe I got the formula wrong or am I misinterpreting something? – vanao veneri Nov 21 '16 at 10:18
• Is the top equation that begins with "S=..." yours or did it come from a text? Our results are the same except you divided by 2. I divided by 4 in order to get the average progenitor IQ weighted by the quantity of offspring produced (2.1+1.9=4 was the total quantity of offspring produced). – sterid Nov 21 '16 at 14:07
• I got it from another post here (biology.stackexchange.com/a/16807/27686). Might be wrong. Your calculation seems logical anyways. – vanao veneri Nov 21 '16 at 14:51
• Meanwhile, I also found a paper here (sciencedirect.com/science/article/pii/S016028961000005X) and there is the following equation $$S=\frac{1}{N}\sum_{i=1}^{N}(IQ_i - \overline {IQ})*\frac{CH_i}{\overline {CH}}$$ Where $CH_i$ is "Children of individual" and $\overline {CH}$ is "Average children in sample". But I do not really understand why we do not simply weigh with the number of offspring per individual as you did. – vanao veneri Nov 21 '16 at 15:25