In Rumbunnies spade tail (A) is dominant to trident tail (a) and Hairy tail (B)

ID: 16601 • Letter: I

Question

In Rumbunnies spade tail (A) is dominant to trident tail (a) and Hairy tail (B) is dominant to naked tail (b). Ziggy a 10th year grad student carried the following cross for his PhD advisor.

Spade/Hairy (AaBb)X Spade/Hairy (AaBb)
He obtained 2400 progeny with the following results:
Spade/Hairy 1192
Spade/naked 608
Trident/Hairy 418
Trident/naked 182
Based on 9:3:3:1 Null hypothesis, Ziggy calculated the expected numbers (Spade/Hairy 1350, Spade/naked 450, trident/Hairy 450, trident/naked 150), and tested for goodness-of-fit Chi-square. He rejected his Null hypothesis, and was at a loss of what's going on.

Using your knowledge of Mendelian genetics, suggest an alternative hypothesis for Ziggy's results (you may want to start by testing for segregation at each locus.) Show all calculations

Explanation / Answer

Spade/Hairy (AaBb)X Spade/Hairy (AaBb)
He obtained 2400 progeny with the following results:
Spade/Hairy 1192
Spade/naked 608
Trident/Hairy 418
Trident/naked 182

Null Null hypothesis,: population in hardyweinberg equilibrium

alternative hypothesis :r population in not hardyweinberg equilibrium

Observed

Expected

o-b

(o-b)^2 /E

Spade/Hairy

1192

1350

158

18.49

Spade/naked

608

450

158

55.47

Trident/Hairy

418

450

2.27

Trident/naked

182

150

6.82

2400

83.05

Based on 9:3:3:1 -------

Spade/Hairy 1350, Spade/naked 450, trident/Hairy 450, trident/naked 150

Chi-square VALUE =SUM OF (O-E) /E

=83.05

Chi-square VALUE IS MORE THAN TABLE VALUE AT 1% AND 5% LEVEL OF SIGNIFICANCE , SO NULL HYPOTHESIS IS REJECTED