Table 7.

A nested hierarchy of inheritance models was successfully fitted with at least two genes controlling the conidiophore architectural phenotype to the counts of progeny phenotypes from three crosses (Table 5). Nine iterations were necessary to achieve the desired error tolerance of 10⁻⁸ with iteratively reweighted least square (IRLS). Recommended models are bolded along with their goodness of fit to the counts of phenotypes in crosses. A null hypothesis (H0) is tested against an alternative (HA) with the chi-squared test having degrees of freedom (df).

Model	Χ2	df	p	X2HA − X2H0	df	p for HA vs. H0	Notes
Full epistatic	0.98	2	0.61	-	-	-	H0 = full epistatic
$\alpha \beta =0$	8.39	3	0.04	8.39 − 0.98 = 7.41	1	0.004	H0 = full epistatic
$\alpha \beta =\alpha =0$	25.31	5	0.001	25.31 − 0.98 = 24.33	2	<0.00001	H0 = full epistatic
$\mathit{\alpha }\mathit{\beta }\mathbf{=}\mathit{\beta }\mathbf{=}\mathbf{0}$	9.96	5	0.08	9.96 − 0.98 = 8.98	2	0.01	H0 = full epistatic
$\beta \gamma =0$	9.32	3	0.02	9.32 − 0.98 = 8.34	1	0.004	H0 = full epistatic
$\alpha \gamma =0$	84.57	3	<0.0001	84.57 − 0.98 = 83.59	1	<0.00001	H0 = full epistatic
$\alpha \beta =\beta \gamma =\alpha \gamma =0$ additive	95.76	4	<0.0001	95.76 − 0.98 = 94.78	3	<0.00001	H0 = full epistatic
environmental	124.46	8	<0.0001	124.46 − 0.98 = 123.48	7	<0.00001	H0 = full epistatic
heritability							H2 = (124.46 − 95.76)/124.46 = 0.23 H0 = environmental model H1 = full additive model