Thanks for your query. (By the way, I guess you are in Brown University, USA, over the vacation from your address?) If you don't understand, just drop me a line and tell me where you have difficulties with my explanation below.

At 03:43 PM 26/05/2000 GMT, you wrote:
>1. On the 1998 exam, part d of question 9, regarding blood types and fertility rates. From looking at the data I believe that there is no evidence to suggest that heterozygous women are twice as fertile, however, I am unclear as to how to calculate that mathematically. It seems to me that heterozygous women being twice as fertile would not proportionally increase the frequency of any one type, however, I am not sure.

I think that is exactly what you were meant to infer. You can check it by doing the calculation: what you have to do is output genotype frequencies for each possible mating, multiplied by their probabilities in the population depending on the genotype frequencies (e.g. LmLm male x LmLm female, LmLm male x LmLn female, LmLn male x LmLm female, and so on). Then you double the fraction expected for the offspring of heterozygous females. Then you add up all the resulting frequencies for each genotype. Then, you will get a total fraction, over all genotypes, that is greater than one, of course; so you have to normalize; i.e. divide all the matings by the total fraction to work out what these genotype frequencies are out of 100%.

This was in the days when we actually worked students through a mating table to get the Hardy-Weinberg results in the second lecture; I think that Kevin used to do this, and I think you still probably do it in the first year. However, it was a difficult part of the question, so don't worry too much about it! I think only about 3 people out of a class of 60 got it right -- but they got very good marks as a result!

>2. 1997 exam, question 9, part e (about the larval fitnesses). Is the proper way to calculate this simple to multiply the predicted H-W frequencies (i.e. p2, 2pq, and q2) times the provided fitnesses? In your lecture on Selection and the Single Gene you find genotype frequencies by dividing by the mean frequency. Do I also need to do that here, and if so, why?

Yes, I think you are right, if I understand you here. See also the theoretical section of our lecture where we treated a simple model of selection; i.e. SELECTION AGAINST A RECESSIVE ALLELE:
http://www.ucl.ac.uk/~ucbhdjm/courses/2007/OneGene/OneGene.html

In that case, the fitness of the recessive genotype aa was 1-s, whereas the fitnesses of all the other genotypes AA and Aa was 1. So after multiplying the genotypic frequencies by the fitnesses, you then have the same problem as in the last case, that your total frequencies don't add up to 1 (In the theory case above, they add up to the mean fitness of:
(p^2).1 + 2pq.1 + (q^2).(1-s), which, after you realize that
p^2 + 2pq + q^2 = 1, ends up being simply 1-sq^2.
(note, p^2 means p-squared).

So in order to find the fraction of something over the total, you have to make the total equal to 1, or 100% or however you want to do it; that is all, that is the reason for dividing by the "mean fitness" or "total fraction" after selection, it is so that you can understand how much the fraction within the population of a particular gene or genotype has changed.

>3. Finally, in my understanding the lectures imply that migration can lead to linkage disequilibrium. I am unclear as to how that can occur.

Ah! Well, if you have a population 1 consisting mainly of AB, and another, population 2, of ab, and they exchange genes, then most of the immigrants will be AB into population 2, and ab into population 1. Since, even after mixing, there is not complete assumption of linkage equilibrium after a single generation, AB and ab gametes will predominate in both populations (in unequal ratios, of course), and the recombinant genotypes, Ab or aB, will be rarer. EVENTUALLY, of course, the two populations will have exchanged so many migrants for so long, that they will reach the same gene frequency, and then there will also probably have been long enough that the linkage disequilibrium will be nearly zero. However, if any selection keeps A & B common in 1 and a & b common in 2, the two populations will continue to maintain the disequilibrium.

Hope this helps! Jim