Notation: D = disease allele at gene d = wild-type allele at gene Let’s provide some information about each MOI. AD: Each of the F1 parents is heterozygous for the D allele. SLR: The observed F2 data are the offspring from an F1 cross where the father is affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That is, the mother’s genotype has one disease allele and one WT allele in it. In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify by gender. Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 Disease, Female 267 WT, Male 285 WT, Female 301 Total 1157 DF p-value
Notation:
D = disease allele at gene
d = wild-type allele at gene
Let’s provide some information about each MOI.
AD: Each of the F1 parents is heterozygous for the D allele.
SLR: The observed F2 data are the offspring from an F1 cross where the father is
affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That
is, the mother’s genotype has one disease allele and one WT allele in it.
In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify
by gender.
Table 1
| O | E | O-E | (O-E)^2 | (O-E)^2 /E | |
| Disease, Male | 304 | ||||
| Disease, Female | 267 | ||||
| WT, Male | 285 | ||||
| WT, Female | 301 | ||||
| Total | 1157 | ||||
| DF | |||||
| p-value |
In Table 2, provide expected proportions for two different MOIs: AD and SLR. The
values in this table are computed using information from the Punnett Square and the
specified MOI. Once the proportions are determined, we can fill in the values in
Table 3, E(xpected) column.
Table 2
| Phenotype | AD proportions | SLR proportions |
| Disease, Male | ||
| Disease, Female | ||
| WT, Male | ||
| WT, Female |
Table 1 twice (make two copies of Table 1), once using
the AD proportions in Table 2 to compute the E(xpected) column, and once using
the SLR proportions in Table 2 to compute the E(xpected) column. Then, follow
through, decide whether either/both/neither of the specified MOIs are consistent
with the observed data, and report your results.
Table 3
| Phenotype | O | E | O-E | (O-E)^2 | (O-E)^2 /E |
| Disease, Male | 304 | =1157 * 0.375 (433.875) | |||
| Disease, Female | 267 | ||||
| WT, Male | 285 | ||||
| WT, Female | 301 | =1157 * 0.125 (144.625) | |||
| Total | 1157 | ||||
| DF | |||||
| p-value |
You must fill in the rest of the cells, compute the Chi-Square statistic, determine the
degrees of freedom (DF), compute the p-value, and determine whether you reject or
do not reject the null hypothesis that you specified the correct MOI.
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