Question

Experiment 2 - Dhybrid Cross: This time, the P generation consisted of a 2.spot male who also had long antennae which is dominant (TTAA) that mated with a no-spot female with short antennae which is recessive (taal. Their F1 offspring all be 2-spot and long antennae and are allowed to mate. The following table shows the actual number of F2 individuals 2-Spot 2-Spot TNO-Spots NO spots Short Ant. Lang Ant. Short Ant F. (observed number) Long Ant. 100 37 56 23 F. (expected ratio) F. (expected number) obs-exp (obs - exp) Cobs - exp) exp What was the hypothesis that you tested with these Chi-square calculations? Do you accept or reject your hypothesis? Why?

Answer 1

In this experiment, a dihybrid cross is performed in which F2 generation will be in 9:3:3:1 ratio. In F2 generation , cross will be in heterozygous parents that will produce 16 progeny. 9 contain both dominant trait, 3 will contain one dominant trait, 3 will contain another dominant trait and 1 contain both recessive trait.

Total observed number= 100+37+56+23 = 216

Expected ratio = 9 : 3: 3 : 1

Expected number= 2-spot and long antennae 216×9/16 = 121.5

2- spot and short antennae 216×3/16= 40.5

No spot and long antennae 216×3/16= 40.5

No spot and short antennae 216×1/16=13.5

Obs-exp = -21.5 -3.5 . 5.5. 9.5

(Obs-exp)2 = 462.25. 12.25 . 30.25 . 90.25

(O-E)2/E = 3.80 0.30 . .74. 6.68

Total = 11.52

Chi value = 11.52

Degree of freedom= 4-1= 3

In this experiment variables are four, hence n-1 means 4-1= 3 will be degree of freedom.

In this experiment, we have used chi square calculation to test hypothesis which states that difference in observed and predicted ratio of dihybrid cross is due to experimental error alone.

In this test, we have reject this hypothesis because in chi square table degree of freedom 3 at 5% level is 7.82 which is lower than the chi value 11.52. To accept this hypothesis, chi value should be lower than the value at 5% level in table. Hence, the difference in ratio is not due to experimental error alone.