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 Map the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. (J. A. Moore. Journal of Heredity 1943, 34:3-7) Cross Parent phenotypes Progeny phenotypes, observed burnsi x burnsi burnsi x pipiens burnsi x pipiens 39 burnsi, 6 pipien:s 23 burnsi, 33 pipiens 196 burnsi, 210 pipiens 2 For each cross, use a chi-square test and the observed numbers of progeny to the expected numbers of progeny ble to evaluate the fit of the Cross 1 Cross 2 Number Number Tools ntinues. Scroll down Do the observed numbers of burnsi and pipiens progeny fit the expected numbers Do the ob x 10 pipiens p rs of burnsi and xpected numbers? O yes yes no no

Explanation / Answer

1) brunsi * brunsi

We got to compare the observed value to expected value

The ratio has to be 3:1 so to get the expected value the total is multiplied by 3/4 and 1/4 fractions respectively. If we observed the table

Calculated X2 = 17.1285

We have to see the row of the table to find out which is the degree of freedom.(N-1, where N stands for number of categories). We have establish probability level, p<0.05 as standard.that is 5% of the column.Now we have to consult a table for critical values of chi square.

Now the calculated value is greater than the tabled value, the hypothesis is wrong or rejected which is the result of chi square value given in number one cross .This conclude brunsi* brunsi it is not an simple Mendelian .

2) Brunsi* pipens

Calculated X2 = 34.380914

The calculated value is greater than the tabled value. This means Hypothesis is rejected and Brunsi* pipens is not simple mendelian cross.

3) brunsi * pipens

Calculated X2 = 154.6436

The calculated value is greater than the table value.This means the hypothesis is rejected and brunsi* pipens does not come under simple mendelian cross.

Phenotypes Observed (O) Expected (E) O-E (O-E)2 (O-E)2/E Brunsi 39 3/4*45=33.75 5.25 27.5625 0.816 Pipens 6 1/4*45=11.25 -5.25 27.6625 16.312 Total 45 45 0 17.1285

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.