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phenotype Observe (O) Expected ratio Expected (E)

PL 284 9/16 214,3125

Pl 21 3/16 71,4375

pL 21 3/16 71,4375

Pl 55 1/16 23,8125

pL 21-71,4375 -50,4375 2543,941406 35,61

pl 55-23,8125 31,1875 972,660 40,85

Expected is from = expected ratio X total progeny observed

Total number of progeny is 381 individuals. Chi-square formula is X2= {(o-e)2}/e Phenotype O-E (O-E)2 (O-E)2 / E ∑(O-E)2 / E

PL 284-214,3125 69,6875 4856,347656 22,67 134,74

Pl 21-71,4375 -50,4375 2543,941406 35,61

DF= phenotype-1 DF= 4-1 = 3

Look to the r= 3 (df) And α= 5% or 0,05 (in 0,950 row) The value is 7,815, because the X2 counting > X2 table So the hypothesis that the alel segragate freely(independently) and all equaly viable are rejected. How about linkage??? Notice That homozygote recessive only can be inherited If Two pl genotype are met. So 55/341= 0,16~

So if probabilitas of ppll is probabilitas of pl times pl, so if the ppll is 0,16 then the pl is the square root of 0,16, then pl is 0,4. Next pl is one of the parental genotype beside PL, so the sum of parental is (PL+pl) is 0,4+0,4= 0,8.... Then the recombinan is 0,2 or 20%.

So the distance is 20 mU.

PL (0,4) Pl (0,1) pL (0,1) pl (0,4)

PL (0,4) PPLL (0,16) PPLl (0,04) PpLL (0,04) PpLl (0,16)

Pl (0,1) PPLl (0,04) PPll (0,01) PpLl (0,01) Ppll (0,04)

pL (0,1) PpLL (0,04) PpLl (0,01) ppLL (0,01) ppLl (0,04)

pl (0,4) PpLl (0,16) Ppll (0,04) ppLl (0,04) Ppll (0,16)

pL 21 0,09 34,29 -13,29 176,6241 5,15

Pl 55 0,16 60,96 -5,96 35,5216 0,42

Expected is from = expected ratio X total progeny observed phenotype Observe (O) Expected ratio Expected (E) O-E (O-E)2 (O-E)2 / E ∑(O-E)2 / E

PL 284 0,66 251,46 32,54 1058,8516 4,21 14,93

Pl 21 0,09 34,29 -13,29 176,6241 5,15

DF= phenotype-1 DF= 4-1 = 3

Look to the r= 3 (df) And α= 5% or 0,05 (in 0,950 row) The value is 7,815, because the X2 counting > X2 table

Still the hypotesis for lingkage is rejected.

If compared, from linked or not, it is more likely to be linked.... but still not well accepted in α 0,05.