Thursday, February 18, 2016

Boosting IQ by 15 points (truncation selection)

It's 2025 and England, having left the EU, has to fend for itself in competitive battle against China. It's not easy: the Chinese are very smart and out-innovate us in key sectors - financial services and high technology.

Something has to be done.

A secret Government task force reports. The problem is our human capital: we're just not smart enough; we can't hack the highly abstract, interconnected and complex realities of the new economy. We have to give our population an IQ boost.

Farmers have known for centuries what to do; animal breeding and genetics professionals have worked out all the details. You identify the traits you want and select that proportion of the population which - if bred together - will deliver offspring with the desired characteristics.

Evolution in action.

'What do we need?' asks the Minister.

Nothing less than a boost in our average population IQ of 15 points, the report suggests, which would bring us up to the level of the the Ashkenazim or India's Brahmin - and better than the Chinese average IQ of 106.

'OK,' says the Minister, 'and how would we get there?'

The report has done the calculations. We'd have to restrict breeding to those people with an IQ greater than 109. Basically, middle-class professionals and up.

The Minister ponders: 'We're doomed.'


The report had an easy-to-read annex which is reproduced below.

State-enforced eugenics has had a deservedly poor press. But voluntary, on-demand child improvement - without coercion - is an easier sell. Putting breeding restrictions to one side, the report also talked about making gene-editing and embryo-selection options widely available, perhaps free on the NHS for the poor.

Perhaps England can hack the 21st century after all.


Truncation selection

As a running example, we take a population of potential parents with mean IQ 100 and standard deviation 15 IQ points; this is broadly the Caucasian English population. We wish to identify a proportion of high-IQ individuals (p%) to breed from, so that their offspring will have an average IQ one standard deviation, σ = 15 IQ points, greater than the current overall parent population.

The related questions of interest:
  1. What is the IQ cut-off threshold, above which we permit high-IQ parents to breed?
  2. What proportion of the overall population parents do we allow to breed?
  3. What is the average IQ of the selected parent-breeders?
We know that the parent-breeder IQ has to be higher than the target IQ in the offspring, because of regression to the mean. This reflects that some of the parents' higher intelligence is due to "luck" - what we euphemistically call 'environmental influences'. In the offspring, this luck goes away and only that portion of intelligence due to (additive) genetics contributes. This is captured by the heritability being less than one.

S is the mean of the selected population minus the population mean, (example: 118.75 - 100 = 18.75, as we shall see).

h2 is the heritability of the trait (example: for IQ we'll assume h2 = 0.8 - estimates vary)

R is the mean of the offspring of those selected minus the population mean, (example: 115 - 100 = 15).

We can now define the relationship between the mean incremental-IQ in the selected parent-breeders and the mean incremental-IQ of their offspring. It's the breeder's equation.
R = h2S.
Knowing R and h2 we can easily work out S. For our running example, we want a future breeding population with mean IQ of 115 (ie R is 15) and the heritability, h2, is 0.8.
So S = 15/0.8 = 18.75.
The mean IQ of our selected parents has to be 118.75.

Connecting proportion allowed to breed (p) with their mean trait-value (S)

This doesn't tell us what proportion of the population is going to be allowed to breed. We need another equation relating S and p, the proportion of the population (on the right-hand side of the bell curve) allowed to breed.

Note that p is the area under the curve on the right of the distribution (see picture above), and S is the mean value of that selected area (in units of σ).

This equation is:
S = σi(p)
where σ is the population standard deviation (15 IQ points) and i(p) is a function we look up in tables, called the 'intensity of selection'. Once we know i(p), we can work backwards in the table to look-up p.

Let's do it.
i(p) = S/σ = 18.75/15 = 1.25. *
Looking up in the tables, p = 26%.

Looking at the selection intensity table below, if we select 26% of parents from the top of the existing intelligence distribution and allow them to breed, the average IQ if their children will be 115, one standard deviation greater than at present.

Note that the average IQ of these parents will be, as we already saw, 118.75.

The IQ cut-off (below which we don't allow anyone to breed) is the truncation point x0. From the table below, it's 0.643 (standard deviations) which equates to an IQ of approximately 110.

The result of all this hard work? Our selected offspring will have an average IQ of 115 - that puts them on a par with the Ashkenazim or Brahmin, and better than the Chinese average of 106.

Job done in one generation!


* We can combine the two equations to eliminate S, so that
R = h2σi(p), or

i(p) = R/(h2σ)   -   (example: i(p) = 15/(0.8 * 15) = 1.25 as above).

Here is the table for i(p).

Further Reading
  1. A slide overview (PDF).
  2. A useful handbook, 'Selection and Genetic Change' (PDF), by Erling Strandberg and Birgitta Malmfors . 
  3. The appendices to the above (PDF) with the maths for the intensity of selection function.