Question

Determine the probable mode of inheritance of each phenotypic trait, the genotypes associated with each phenotype and the ratios of genotypes and phenotypes expected in crosses if your model is correct. You will confirm this by using a chi-square test. Wild Population - 100 mice (1 to 100) Both parents collected from the wild Body Size FM Total Large 14 27 41 Medium 110 11 Small 17 31 48 Line 1 - 160 mice (101 to 260) Parents: (Body Size) M: (Vial O, #1) Small F: (Vial O, #2) Large Body Size FM Total Large 076 76 Medium 840 84 Small 0 0 0 Line 2 - 104 mice (261 to 364) Parents: (Body Size) M: (Vial 1, #102) Large F: (Vial 1, #101) Medium Body Size F M Total Large 25 22 47 Medium 270 27 Small 0 30 30 Line 3 - 180 mice (365 to 544) Parents: (Body Size) M: (Vial 2, #261) Small F: (Vial 2, #263) Large Body Size FM Total Large 092 92 Medium 88 0 88 Small 00d

Answer 1

Line 1 2 and 3 is sex linked codominant mode of inheritance.

But in case of wild population it cannot be sex linked as all the phenotypes are present. So it must be codominant autosomal trait.

A A. Large xay x xay. & Expected ratio V Av dediler. = go male e so Romale x^x large 322420.4. 16 As 5% value of x is less that tabulated value at probability level, the model is correct ма: Клу xxAx4 1 1 1 1 - 1 - XAXA XAxa XAy xay diun Targe Targe small, Phenotypecto = 26 large female 826 large male - 8 26 medium female & 30 small male Леко ури Page 2 16 as it has to the 320< 123 26,2 in As value of ne is ress than tabulated values, the model is correct. Lune 3: XxAxx"Y wala 90 medium x² 4 90 x² is 4 a 84 a 0.089 90 4045 less model to correct. cale

Aa X Aa

AA Aa Aa aa

So ratio should be 1:2:1 ie. 25 large, 50 medium and 25 small.

So chi sq= 61.82 which is larger than tabulated value. So the ratio is significantly different.