I have determined that at locus 1, the mode of inheritance is sex linked and at
ID: 276656 • Letter: I
Question
I have determined that at locus 1, the mode of inheritance is sex linked and at locus 2, the mode of inheritance is autosomal recessive.
Could someone help me determine the expected number of individuals that will have the following phenotypes, and fill out the E column.
Chi Square test:
Gender
Two locus phenotype
Observed (o)
Expected (E)
(o-E)^2 /E
Female
WT/ Disease
0
Disease/Disease
0
Disease/ WT
255
WT/WT
258
Male
Disease/WT
252
WT/disease
0
WT/WT
260
Disease/Disease
0
Total
1025
DF=
P value=
Gender
Two locus phenotype
Observed (o)
Expected (E)
(o-E)^2 /E
Female
WT/ Disease
0
Disease/Disease
0
Disease/ WT
255
WT/WT
258
Male
Disease/WT
252
WT/disease
0
WT/WT
260
Disease/Disease
0
Total
1025
DF=
P value=
Explanation / Answer
Expected value ( Total of observed values/total number):
( 1025/8 )
Observed - expected
( Observed - Expected )2
( Observed - Expected )2/exp
Total = 1,025.284
DF ( degree of freedom ) = N-1
N=8
DF = N-1 = 8-1 = 7
P value = It depends upon hypotheseis
HO: u = 0
Ha: u is not equal to zero
P value depends on the Z table
You can see the result of p value on the Z table. You have a degree of freedom 7 and at the point of 7 you can check the p value.Expected value ( Total of observed values/total number):
( 1025/8 )
Observed - expected
( Observed - Expected )2
( Observed - Expected )2/exp
Total = 1,025.284
DF ( degree of freedom ) = N-1
N=8
DF = N-1 = 8-1 = 7
P value = It depends upon hypotheseis
HO: u = 0
Ha: u is not equal to zero
P value depends on the Z table
You can see the result of p value on the Z table. You have a degree of freedom 7 and at the point of 7 you can check the p value.