Alkapt-ascertainTbl

# Alkapt-ascertainTbl - Obs minus expected Affected = 61-56.9...

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EXAMPLE: Make a table OF DISTRIBUTION OF Alkaptonuria in FAMILIES - Ignore families of n=1. They are uninformative =100% Family (sibship) size n Number of families x total offspring nx q' expected affected nxq' Observed affected 1 5 5 1 5 5 2 8 16 .5714 9.14 10 3 5 15 .4324 6.48 8 4 2 8 .3657 2.93 4 5 3 15 .3278 4.92 4 6 3 18 .3041 5.47 8 7 2 14 .2885 4.04 5 8 3 24 .2778 6.67 5 9 1 9 .2703 2.43 4 10 1 10 .2649 2.65 2 11 3 33 .2610 8.61 7 12 0 - .2582 na 0 13 0 - na 0 14 1 14 3.56 4 TOTAL 37 176 56.9 61 61 observed is very close to the 56.9 expected SO we conclude it is a recessive Mendelian trait NOW TEST THE RESULT WITH X 2 TEST Sum (o-e) 2 /e = X 2 Must remove n = 1 sibships because they are uninformative; hence 61, 56.9 and 176 all lack the five offspring in one children families Normals = total – affected, = 176-61 = 115obs norm; 176 – 56.9 = 119.1exp normal
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Unformatted text preview: Obs minus expected Affected = 61-56.9 = 4.1 Obs minus expected normal = 115-119.1 = 4.1 (61-56.9) 2 /56.9 = (4.1) 2 /56.9 = 0.295 (106-119.1) 2 /119.1 = (4.1) 2 /119.1 = 0.141 X 2 = 0.436 1df p ~= 0.5 Conclusion: If the hypothesis is true (that alkaptonuria = simple recessive and we have correctly accounted for bias) then the difference between these data and the expected data would occur ~50% of the time. This is sufficiently likely for us to conclude that the "null hypothesis" is correct, namely that there is no significant difference between the observed and expected numbers....
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## This note was uploaded on 10/27/2009 for the course BIO SCI 05580 taught by Professor Marsh during the Fall '09 term at UC Irvine.

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