J. A. Moore investigated the inheritance of spotting patterns in leopard frogs.

Question

J. A. Moore investigated the inheritance of spotting patterns in leopard frogs. The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype k. Moore carried out the cross of burnsi x burnsi, which produced 39 burnsi and 6 pipiens offspring. (J. A. Moore. Journal of Heredity 1943, 34:3-7). spots on its Cross Parental phenotypes Progeny phenotypes, observed burnsi x burnsi 39 burnsi, 6 pipiens burnsi x pipiens 23 burnsi, 33 pipiens burnsi x pipiens 196 burnsi, 210 pipiens g Plus for Pier als: Concepts a r each cross, use a chi-square test and the chi-square critical values table to evaluate the fit of the erved numbers of progeny to the expected numbers of progeny Cross 1 Cross 1 Number Do the observed numbers of burnsi and pipiens progeny fit the expected numbers? Do the observed numbers of burnsi and pipiens progeny fit the expected numbers? O yes O yes O no no Hint terms of use us help ut us careers privacy policy

Explanation / Answer

Answer:

Burnsi (PP/Pp) dominant over Pipiens (pp)

1. Cross bet ween Burnsi X  Burnsi

Note. Both are Hetrozygous condition, then only we can observe two diffrent phenotypes (Burnsi and Pipiens)

Pp X Pp

1PP: 2Pp:1pp= 3 Burnsi: 1 Pipiens ; total = 45 progeny

Degree of freedom (n-1)= 2-1=1

X2 Critical value from tabular is 3.84 at 5% error. So X2 total < X2 critical; 3.27< 3.84.

Yes: Observed number of Burnsi and Pipiens fit in expected number and given hypothesis is accepted.

2. Cross bet ween Burnsi X  Pipiens

Note. Burnsi is Hetrozygous condition, then only we can observe two diffrent phenotypes (Burnsi and Pipiens)

Pp X pp

1Pp:1pp= 1 Burnsi: 1 Pipiens ; total = 56 progeny

Degree of freedom (n-1)= 2-1=1

X2 Critical value from tabular is 3.84 at 5% error. So X2 total < X2 critical; 1.78< 3.84.

Yes: Observed number of Burnsi and Pipiens fit in expected number and given hypothesis is accepted.

3. Cross bet ween Burnsi X  Pipiens

Note. Burnsi is Hetrozygous condition, then only we can observe two diffrent phenotypes (Burnsi and Pipiens)

Pp X pp

1Pp:1pp= 1 Burnsi: 1 Pipiens ; total = 406 progeny

Degree of freedom (n-1)= 2-1=1

X2 Critical value from tabular is 3.84 at 5% error. So X2 total < X2 critical; 0.48< 3.84.

Yes: Observed number of Burnsi and Pipiens fit in expected number and given hypothesis is accepted.

S.NO Observe (o) Expexted (e) o-e (o-e)2 (o-e)2/e X2 Value X2 Burnsi 39 33.75 5.25 27.6 0.82 0.82 X2 Pipiens 6 11.25 -5.25 27.6 2.45 2.45 X2 Total 3.27

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