Wednesday, March 07, 2007
lwka's excellent post below mentions (among other things) the low reported IQ in African countries.
When this came up on a previous occasion the question was raised whether mean IQ's of 70 or even lower for Africans were compatible with a 'black' mean IQ of 85 in the United States, after taking account of the mixed African and European ancestry of American blacks. At that time I did a few sums and started writing a post, but never quite finished it. But I have dusted it off and posted it below the fold, as it still seems relevant. First, some assumptions: - for the sake of this discussion, I assume that all group differences are wholly genetic in origin. Of course, I don't believe this, and I'm not sure that anyone does. - I assume that the observed mean IQ of white Americans, on current (2000-ish) norms, is 100, and that of 'black' Americans is 85 - I started by assuming that, on average, 'black' Americans have 20% white ancestry. However, having done all the calculations on this basis and written up the results, I found that a figure of 25% is also sometimes used (e.g. by Jensen). This seems rather high, (see the figures in Sandra Scarr's Race, Social Class and Individual Differences in IQ), but I have repeated the calculations on this basis as an alternative. Rather than present two sets of workings, I have just inserted the key alternative results in square brackets. Whether on the 20% or 25% basis, I assume that the white ancestry of 'blacks' is genetically representative of the white American population.Needless to say, all of these assumptions could be discussed at length. With these assumptions, if all genetic effects are strictly additive, with no dominance or epistasis, then the mean IQ of an unmixed 'black' population would be X, where .8X + .2(100) = 85, which gives the value X = 81.25 [or 80 on the alternative 25% basis]. This is the easy bit! But it is unlikely that all the genes affecting IQ are purely additive. To consider the possible effects of dominance, let us assume that all genes affecting IQ have two alleles, H (High) and L (Low), with H dominant. To take the most extreme assumption for gene frequencies, maximising the effect of dominance in differentiating a 'pure' black population from a black-white mixture, let us assume that the white population is entirely homozygous for the dominant H, while the ancestral black population is entirely homozygous for the recessive L. The US mixed 'black' population would therefore have a gene frequency at each locus of 80% L and 20% H. In Hardy-Weinberg equilibrium this gives genotype frequencies of .64LL, .32HL, and .04HH. The phenotypic value of each locus in the US 'black' population would therefore be .64X + (.32 + .04)100 = 85, where X is the unknown phenotypic value of the 'pure' black population. This gives X = 76.56 [or 73.2 on the 25% basis]. This is not a realistic assumption for gene frequencies, since if it applied to all loci the 'pure' populations would be genetically uniform, which is obviously not the case. To consider a more plausible model for gene frequencies, suppose the frequency of H is 60% in the white population and 30% in a pure black population, so that the frequency of L is 40% in the white population and 70% in a pure black population (since the frequency of H + L must be 100% for each population.) For the US 'black' population the frequencies are therefore (.8 x 30%) + (.2 x 60%) = 36% for H, and 64% for L. The Hardy-Weinberg genotype frequencies for the white population are 36%HH, 48%HL, and 16%LL and for the US 'black' population (approx) 13%HH, 46%HL, and 41%LL. If we designate the phenotypic value of HH and HL as X (not to be confused with the previous X!) and of LL as Y, we get two equations as follows: (.36 + .48)X + .16Y = 100 (.13 + .46)X + .41Y = 85 Solving these equations gives (approx) X = 109.6 and Y = 49.6 [or 110.4 and 45.4 on the 25% basis]. In a 'pure' black population with gene frequencies of 30% H and 70% L, and Hardy-Weinberg genotype frequencies, these results give a mean phenotypic value for the population of almost exactly 80 [78.5 on the 25% basis], as compared with 81.25 [80 on the 25% basis] if there is no dominance. It will be seen that in this model the effect of dominance in differentiating 'pure' from 'mixed' black population is not very great. These are in principle the values for a single locus. In the absence of epistasis, the mean value for the genome as a whole is simply the mean value of all relevant loci. Epistasis, by definition, means that the phenotypic value of the genome varies according to the particular combination of genes at different loci. Essentially the only limit on possible epistatic effects is the ingenuity of the model-maker. As a general rule, epistasis is less important than dominance, so my gut feeling is that epistasis, on any plausible model, is unlikely to make much difference to the conclusions above. If anyone disagrees, let them set out a plausible model in which it does make a difference. Note that if high white IQ is due to favourable epistatic gene combinations, these are very likely to be broken up in a 'black' population with only 20% or 25% white ancestry, so that in this respect the difference between a 'pure' black population and the US 'black' population would probably be small. Departures from Hardy-Weinberg equilibrium would also be a complication. Assortative mating within the US 'black' community for traits deriving from white ancestry, such as skin colour or IQ (assuming that this is genetic), would tend to increase the proportion of homozygotes above the Hardy-Weinberg expected levels. This would reduce the overall phenotypic contribution of white ancestry, since the same number of dominant H genes would be distributed among fewer individuals than in Hardy-Weinberg equilibrium. In effect, one of the dominant genes is 'wasted' in a homozygote. On the other hand, first generation black-white hybrids would have a higher proportion of heterozygotes than the rest of the 'black' population. One prediction of a purely genetic model assuming dominance for high-IQ genes is that the IQ of first-generation mixed-race individuals would be closer to the white mean IQ. But this would presumably affect only a small proportion of the 'black' population as a whole. I haven't bothered calculating the effects on the assumption that the genes for low IQ are dominant. Obviously in that case the 'mixed' black population would be closer to the 'pure' black population.The general conclusion is, I think, that on hereditarian assumptions about the nature of the black-white IQ difference in the US, the genetic IQ of a 'pure' black population cannot be much below 75 and is more likely around 80. (Note that this is by present-day norms. Due to the Flynn Effect, a present-day score of 80 would have corresponded to about 100 when IQ tests were first established.) This is higher than the mean IQ (often below 70) sometimes reported for black African countries from which the black ancestry of 'black' Americans is derived. Of course, one could in principle argue that the Africans who were taken to the US were above average in genetic IQ (relative to other Africans), but this is hardly likely. Alternatively, one could argue that 'black' Americans have improved their genetic IQ, and/or that African genetic IQ has deteriorated, in the 250 years or so since the slaves arrived in America. Neither hypothesis seems plausible. It would therefore be difficult to consistently maintain a purely hereditarian view of both the black-white difference in the US and the low measured IQ of black Africans. I am not sure that anyone has actually take this position, but it may still be worth pointing out the difficulties. Labels: IQ |