Just need help because under observed I put 81 and 72 then to calculate the expe
ID: 40384 • Letter: J
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Just need help because under observed I put 81 and 72 then to calculate the expected is where I'm lost! crossed a worm homozygous for the dpy-17 e164 allele with a male worm heterozygous for dpy-17 e164. She scored all of the developing eggs/larva for the Dpy phenotype and found 72 with the Dpy phenotype and 81 that were wild type. Do a Chi square test on phenotypic data to determine if your data supports the hypothesis that the mutation segregates as a single-gene mutation. Show your work in a table and include a. Null hypothesis b. X^2 value c. dof d. p value e. ConclusionExplanation / Answer
null hypothesis ;
it show s that there will be no relation between the two phenotypes ,it depends on P value , dispproving this hypothesis leads to the there is a relation between the molecules .
chi square analysis ;
it is the good ness of fit value /, where the expected and observed values are bought to the summation .An hypothesis is developed thatpredicts how a set of observations will fall into expected These counts are compared with the experimental data that is observed Allowingfor the sample size, the differences among theobserved and expected results are reduced to asingle number, the chi-square value.
O IS THE OBSERVED VALUVE =71
E IS THE EXPECTED VALUVE =81
SUM IS THE DEVIATION .
= SUM { 71-81}2 /81 } =3.16/ 81 = 0.039 IS THE CHI SQUARE VALUE .
dof value
is the degree of freedom , the number of phenotypes taken into account , give n -1 that is 1-1 =0 , so the dof value is zero .
the p value is 1.00
conclusion ; this above experiment do not support the null hypothesis . so there is a single gene mutation
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