The Analysis

	$H_{1}$:	Half of the offspring from heterozygous mothers are also heterozygous.
		Under $H_1$
		$$\begin{eqnarray*} \gamma_{21} &=& (1/2)(F_{2}/F_{0})(C_{21}/(C_{21}+C_{23})) \... ... \gamma_{23} &=& (1/2)(F_{2}/F_{0})(C_{23}/(C_{21}+C_{23})) \end{eqnarray*}$$
		Under $H_1$, $\gamma_{22}$ can be predicted just from the frequency of heterozygous mothers in the sample. Thus, only 9 parameters are needed to describe the data under $H_1$. Since 10 are required if we reject $H_1$ we can use a likelihood ratio test with one degree of freedom to see whether the above estimates provide an adequate description of the data.

If $H_1$ is rejected, we can conclude that there is either gametic selection or segregation distortion in $A_{1}A_{2}$ females.

$H_{2}$:

The frequency of transmitted male gametes is independent of the mother's genotype.

Under $H_2$

$$\begin{eqnarray*} p_{m} &=& (C_{11} + C_{21} + C_{32})/(F_{0} - C_{22}) \\ q_{m} &=& (C_{12} + C_{23} + C_{33})/(F_{0} - C_{22}) \end{eqnarray*}$$

The expected frequency of the various mother-offspring combinations is

	$A_{1}A_{1}$	$A_{1}A_{2}$	$A_{2}A_{2}$
$A_{1}A_{1}$	$\phi_{1}p_{m}$	$\phi_{1}q_{m}$	--
$A_{1}A_{2}$	$(1/2)\phi_{2}p_{m}$	$(1/2)\phi_{2}$	$(1/2)\phi_{2}q_{m}$
$A_{2}A_{2}$	--	$\phi_{3}p_{m}$	$\phi_{3}q_{m}$

where $\phi_{i} = F_{i}/F_{0}$. Under $H_2$ only the female genotype frequencies and the male gamete frequencies are needed to describe the mother-offspring data. That's a total of $2+1+2+2=7$ frequencies needed to describe all of the data. Since $H_1$ needed 9, that gives us 2 degrees of freedom for our likelihood ratio test of $H_2$ given $H_1$.

If $H_2$ is rejected, we can conclude that there is some form of non-random mating in the breeding population or female-specific selection of male gametes.

	$H_{3}$:	The frequency of the transmitted male gametes is equal to the allele frequency in adult males.
		Under $H_3$ the maximum likelihood estimates for $p_m$ and $q_m$ cannot be found explicitly, they are a complicated function of $p_m$ and $q_m$ as defined under $H_{2}$ and of $M_{1}$, $M_{2}$, and $M_{3}$. Under $H_3$, however, we no longer need to account separately for the gamete frequency in males, so a total of $2+2+2=6$ frequencies is needed to describe the data. Since $H_2$ needed 7, that gives us 1 degree of freedom for our likelihood ratio test of $H_3$ given $H_2$.

If $H_3$ is rejected, we can conclude either that males differ in their ability to attract mates (i.e., there is sexual selection) or that male gametes differ in their ability to accomplish fertilization (e.g., sperm competition), or that there is segregation distortion in $A_{1}A_{2}$ males.

	$H_{4}$:	The genotype frequencies of reproductive females are the same as those of ``sterile'' females.
		Under $H_4$ the maximum likelihood estimates for the genotype frequencies in females are
		$$\phi_{i} = (F_{i}+S_{i})/(F_{0}+S_{0})$$
		Under $H_4$ we no longer need to account separately for the genotype frequencies in ``sterile'' females, so a total of $2+2=4$ frequencies is needed to describe the data. Since $H_3$ needed 6, that gives us 2 degrees of freedom for our likelihood ratio test of $H_4$ given $H_3$.

If $H_4$ is rejected, we can conclude that females differ in their ability to reproduce successfully.

	$H_{5}$:	The genotype frequencies of adult females and adult males are equal.
		Under $H_5$ the maximum likelihood estimates for the adult genotype frequencies can not be found explicitly. Instead, they are a complicated function of almost every piece of information that we have. Under $H_5$, however, we no longer need to account separately for the genotype frequencies in females and males, so a total of $2$ frequencies is needed to describe the data. Since $H_4$ needed 4, that gives us 2 degrees of freedom for our likelihood ratio test of $H_5$ given $H_4$.

If $H_5$ is rejected we can conclude that the relative viabilities of the genotypes are different in the two sexes. (We have assumed implicitly throughout that the locus under study is an autosomal locus. Notice that rejection of $H_5$ is consistent with no selection in one sex.)

$H_{6}$:

The genotype frequencies in the adult population are equal to those of the zygote population.

Under $H_6$ the maximum-likelihood estimator for the allele frequency in the population is

$$p = {((C_{11}+C_{21}+C_{32})+2(F_{1}+S_{1}+M_{1})+(F_{2}+S_{2}+M_{2})) \over ((F_{0}-C_{21})+F_{0}+S_{0}+M_{0})}$$

Under $H_6$ the genotype frequencies in our original table can be summarized as follows:

Mother	$A_{1}A_{1}$	$A_{1}A_{2}$	$A_{2}A_{2}$	$\sum$	``Sterile'' Females	Males
$A_{1}A_{1}$	$p^3$	$p^{2}q$	0	$p^2$	$p^2$	$p^2$
$A_{1}A_{2}$	$p^{2}q$	$pq$	$pq^2$	$2pq$	$2pq$	$2pq$
$A_{2}A_{2}$	0	$pq^2$	$q^3$	$q^2$	$q^2$	$q^2$

In short, under $H_6$ only one parameter, the allele frequency, is required to describe the entire data set. Since under $H_5$ needed two parameters, our likelihood ratio test of $H_6$ given $H_5$ will have one degree of freedom.

If $H_6$ is rejected, we can conclude that genotypes differ in their probability of survival from zygote to adult, i.e., that there is viability selection. If $H_1$-$H_6$ are accepted, we have no evidence that selection is happening at any stage of the life cycle at this locus and no evidence of non-random mating with respect to genotype at this locus.