Question

You are studying the genetics of the fruit fly. Body color (brown or black) is controlled by the b locus and wing size (long or short) is controlled by the vg locus.

In cross 1, you mate two true-breeding strains and obtain the following progeny:

cross 1: brown, short-winged male X black, long-winged female

F1: all brown, long-winged flies

a) Which phenotypes are dominant and which phenotypes are recessive?

i) The recessive trait for body color is _____________________.

ii) The recessive trait for wing size is ______________________.

b)                            In cross 2, you then mate many pairs of siblings from the F1 generation of cross 1:

cross 2: F1 sibling male X F1 sibling female

What genotypic and phenotypic ratios would you expect in the progeny (F2) of these crosses if independent assortment occurs?

The compiled progeny results from the cross 2 mating pairs are presented below:

F2:                  Phenotype                                          # observed

                        Brown, large-winged                         512

                        Black, large-winged                           244

                        Brown, short-winged                         246

                        Black, short-winged                           9

c) From this data:

• Calculate a Chi-square value for these data.             χ²=   

• How many degrees of freedom are there for this cross?  df =

• What is the P value?                                                  P =  

• Do you reject or accept the Mendel's hypothesis?       

Answer 1

Ans1)a)brown color and long wing is dominant phenotype as all F1 flies of true mating have this phenotype.

i) recessive trait for body colour is black colour

ii) recessive trait for wing size is short wing

Ansb) mating betweeb two F1 flies if independent assortment occur then F2 phenotypic ratio is 9:3:3:1

And F2 genotypic ratio is 1:2:1:2:4:2:1:2:1

Explaination:

C)

	Expected frequency	observed frequency	O-E	(O-E)2	(O-E)2/E
Brown long wing	9/16=0.56	512/1014=0.51	-0.05	0.0025	.0045
Black large wing	3/16=0.19	244/1014=0.241	0.051	0.0026	0.0137
Brown short wing	3/16=0.19	246/1014=0.242	0.052	0.0027	0.0142
Black short wing	1/16=0.063	9/1014=0.01	-0.053	0.0028	0.045

i) Chi square value=X²=

(O-E) ²/E=0.0045+0.0137+0.0142+0.045=0.0774

ii) Degree of freedom=no of phenotype-1=4-1=3

iii)table p value at α=0.05 for df =3 is 7.82

The final step is to apply the value generated to a chi-squared distribution table to determine if results are statistically significant

• A value is considered significant if there is less than a 5% probability (p < 0.05) the results are attributable to chance

When df = 3, a value of greater than 7.815 is required for results to be considered statistically significant (p < 0.05)

• Calculated chi square value is  0.077 which is less than table p value so results are caused by chance
• Hence, the difference between observed and expected frequencies are not statistically significant

As results are not statistically significant, the alternative hypothesis is rejected and the null hypothesis accepted:

• Null hypothesis (H₀): There is no significant difference between observed and expected frequencies (genes are unlinked) ie mendel hypothesis