DesignLayou ou predicted a certain ratio of phenotypes in the section above. Use

ID: 207942 • Letter: D

Question

DesignLayou ou predicted a certain ratio of phenotypes in the section above. Use this prediction to work out what numbers of each type of fly you expected in your cross. Compare this to the observed number, using the x test. You may use your own results or those of the class. First construct a Null hypothesis': That... O E)P/E Class Observed (Expected (E) O EP (normal wing, normal eyel normal wing. scarlet eye vestigial wing, normal cye vestigial wing scariet cye Total are calculated from the observed total ·The expected numbers of the different phenotypes and the ratios you predicted above, in the Punnet square (see prac. Notes). Degrees of freedom (g) (number of different classes-1): Critical value of 2efor a probability of 5% and the gtaboe: Accept/reject (delete as appropriate) null hypothesis because: What do you conclude from the y2 test? 120 marks) Further work Suggest other experiments you would like to do, to confirm your conclusions [10 marks) 11 2122 Words

Explanation / Answer

The ratio you got in the first part of the exercise is 9:3:3:1.

Now you need to convert this probabilities in numbers, so you can compare them to the observed from the class.

We do this first, getting the probabilities and multiplying the total.

Total: 1396+486+315+93 = 2290

9/16 : 0.5625 x total (2290)=1288

3/16 : 0.1875 x total (2290)=429

1/16 : 0.0625 x total (2290)=143

This are our expected numbers of individual of each type of phenotype.

Now we use the Chi-square test: 2 in order to compare if the observed ones adjust to the ratio you encountered.

Null hypothesis: Observed phenotypes are not the expected ones by the predicted ratios found in the section above.

Chi-square test: 2

2 = (observed individuals – expected individuals) 2

expected individuals

The table is there to help you solve this equation:

class

Observed

Expected

(O-E)2

(O-E)2/E

1396

1288

11,664

9.06

Nw / Se

486

429

3,249

7.57

Vw/Ne

315

429

12,996

30.3

Vw/Se

143

2,500

17.48

2 = 9.06 + 7.57 + 30.3 + 17.48 = 64,41

The 5% significance level for 1 degree of freedom is 3.84

64,41 > 3.84

This means we reject Ho. The observed phenotypes adjust to the expected ratios you get with the punnet square!