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I've been working a little with M. G. Bulmer's Principles of Statistics (Dover, 1979) and cannot see how to properly compute half of one question. This should be a basic probability computation, but I am not getting the same answer as the book's suggestion.

Here is the question (Chapter 2, Problem 2.4): If the three genotypes (i.e., AA, Aa, aa) are all distinguishable, find the probabilities that a) a pair of siblings and b) a pair of unrelated individuals will appear the same when p=q=.5.

The book offers .59 for a) and .38 for b). I can get b correctly, but I don't understand how to properly compute part a).

This does not seem like a complicated exercise!

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2 Answers 2

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I got .59. In response to WYSIWYG:

(Note: I'm using A and a for the two alleles. My copy of Bulmer uses B for the second, which makes a little more sense if the alleles are codominant.)

a.) The AA x aa cross should be twice as common as the AA x AA and aa x aa crosses, just as HT is more common than HH or TT when we toss a pair of identical coins. In particular, this cross should occur with frequency 2*(.25*.25)= .125. Same goes for AA x Aa, AA x aa, and Aa x aa.

b.) In an Aa x Aa cross, there is a 3/8 probability that two offspring will have the same genotype. I reasoned as follows:

  • There is a 0.5 probability that the first offspring will be heterozygous (Aa). In that case, there is a 0.5 probability that the second offspring will also be heterozygous.

  • There is a 0.25 probability that the first offspring will be homozygous AA: In that case there is a 0.25 probability that the second offspring will also be AA. Likewise for aa.

  • Summing these: (1/2)^2 + (1/4)^2 +(1/4)^2 = 1/4 + 2/16 = 3/8.

I believe that resolves the issue. Apologies if I've missed something.

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  • $\begingroup$ this seems to be right. Thanks for correcting. $\endgroup$
    – WYSIWYG
    Apr 30, 2014 at 7:31
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Are you sure it is 0.59 because I am getting 0.5625 for part (a) and this is how I computed.

To check genetic similarity of siblings we first have to pick up mating pairs. Six mating pairs possible with following offspring probabilities (choice of first mate x choice of second mate x same offspring genotype freq).

  1. AA x AA : will always give same type of offspring: prob = 0.25x0.25x1
  2. aa x aa : will always give same type of offspring: prob = 0.25x0.25x1
  3. AA x aa : will always give same type of offspring: prob = 0.25x0.25x1
  4. Aa x Aa : will give Aa in half the number of times: prob = 0.5x0.5x0.5
  5. Aa x AA : will give two types of similar offsprings, half the number of times - AA and Aa: prob = 0.5x0.25x0.5x2
  6. Aa x aa : will give two types of similar offsprings, half the number of times - aa and Aa: prob = 0.5x0.25x0.5x2

Total probability = 3x(0.5x0.5x0.5) + 3x(0.25x0.25) = 0.375+0.1875 = 0.5625

Assumptions:

  • Population is infinite
  • Population in equilibrium
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    $\begingroup$ For what it's worth, I repeated this calculation in my own peculiar way and also got 0.5625 $\endgroup$
    – Alan Boyd
    Apr 27, 2013 at 8:50
  • $\begingroup$ @AlanBoyd: can you tell me how you approached it ? $\endgroup$
    – WYSIWYG
    Apr 27, 2013 at 8:55
  • $\begingroup$ By peculiar I meant I did it in a spreadsheet - probably it is exactly the same logic as your calculation. See dropbox.com/s/k46xhff3x0p1oc7/HW2png.png $\endgroup$
    – Alan Boyd
    Apr 27, 2013 at 11:22
  • $\begingroup$ Thanks to you both for your comments. .5625 is what I got as when I did this, too. It may simply be an error on the solutions page, but I couldn't rule out that I was making an error. $\endgroup$ Apr 27, 2013 at 13:00

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