ruilfies In fruit flies genes pr and vg code alleles are as

ruilfies In fruit flies, genes pr and vg code alleles are as follows for the prickly\" and vestigial traits. The wing pr prickly back pr* smooth back (no prickly hairs) vg a normal wing vg vestigial (smooth wing) You perform a test cross between a fly with genotype pr pr vg vg. and a fly with genotype pr pr vg yields. 1) Draw a dihybrid Punnett square predicting the progeny of this cross. State all predicted phenotypic outcomes and the expected proportion of each. 291, Pr\", V9 va .25. 2) Below are the actual phenotypic outcomes from the testcross. Observed Phenotype Expected 874 Prickly, normal wing 906 Smooth back, vestigial wing 112 Prickly back, vestigial wing t5 98 Smooth back, normal wing Generate a null hypothesis that you can test to see if the pr and vg alleles are linked (located the same chromosome). Write your null hypothesis below. Cross with one anathur and see the outuem 3) Fill in the \"expected\" column above with the number of each phenotype you would obtaim your null hypothesis were correct.

Solution

A dihybrid Punett square is a square which shows only two characters(traits) of a genotype.

Both the parents taking part in this cross are heterozygous means contain two alleles of a gene in one cell. Here complete dominance is exhibited by one allele of each trait. In this case probability of phenotypic is equal to the genotypic probability. The phenotypic ratio will be 2:4:2:2:2:2:2. The phenotypes are as follow:

a. prickly back-2

b. prickly back and normal wing -4

c. either prickly back or smooth back-2

d. smooth back and normal wing-2

e. normal wing-2

f. smooth back and vestigial wing-2

g. either normal wing or vestigial wing-2

2. Chi square analysis is carried out if there lies a possibility of genes found on same chromosome due to which they are unable to get assorted indepently.

  

	Expected number	Observed number	(observed-expected)2 / expected