I need help in table 2 & 3 please Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 433.875 129.875 16789.68 38.69 Disease, Female 267 433.875 166.875 26855.01 61.89 WT, Male 285 144.625 140.375 19705.14 136.25 WT, Female 301 144.625 156.375 24453.15 156.37 Total 1157 1157 393.20 DF 3 p-value 7.84 Expected progencies as per SLR MOI’s = 1:1 for both male and female Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 289.25 14.75 217.56 0.75 Disease, Female 267 289.25 22.25 495.06 1.85 WT, Male 285 289.25 4.25 18.06 0.06 WT, Female 301 289.25 11.75 138.06 0.46 Total 1157 1157 3.12 DF 3 p-value 7.84 Progenies are following SR MOI’s that parents have genotypes and progenies in ratio 1:1 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.
I need help in table 2 & 3 please
Table 1
|
|
O |
E |
O-E |
(O-E)^2 |
(O-E)^2 /E |
|
Disease, Male |
304 |
433.875 |
129.875 |
16789.68 |
38.69 |
|
Disease, Female |
267 |
433.875 |
166.875 |
26855.01 |
61.89 |
|
WT, Male |
285 |
144.625 |
140.375 |
19705.14 |
136.25 |
|
WT, Female |
301 |
144.625 |
156.375 |
24453.15 |
156.37 |
|
Total |
1157 |
1157 |
|
|
393.20 |
|
|
|
|
|
DF |
3 |
|
|
|
|
|
p-value |
7.84 |
Expected progencies as per SLR MOI’s = 1:1 for both male and female
|
Phenotype |
O |
E |
O-E |
(O-E)^2 |
(O-E)^2 /E |
|
Disease, Male |
304 |
289.25 |
14.75 |
217.56 |
0.75 |
|
Disease, Female |
267 |
289.25 |
22.25 |
495.06 |
1.85 |
|
WT, Male |
285 |
289.25 |
4.25 |
18.06 |
0.06 |
|
WT, Female |
301 |
289.25 |
11.75 |
138.06 |
0.46 |
|
Total |
1157 |
1157 |
3.12 |
||
|
|
|
|
|
DF |
3 |
|
|
|
|
|
p-value |
7.84 |
Progenies are following SR MOI’s that parents have genotypes and progenies in ratio 1:1
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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