A true breeding tall and red flowered hibiscus plant is mate

A true breeding tall and red flowered hibiscus plant is mated to a true breeding short and plant. The F_1 plants are self fertilized yielding the following F_2 data. 1310 Tall Red 420 Tall White 480 Short Red 150 short while what phenotype ration (hypothesis) does the f_2 data approximate? Use chi square analysis to test if the observed numbers in the F_2 data is consistent with your hypothesis. Interpret the chi square results Give the genotypes of the P and F_1 generations Draw a branch diagram showing the expected proportion of each phenotypic class for the F2 data above.

Solution

a). Assume that the gene coding for tall flower is A and its recessive allele a codes for short flower; A is dominant over a. Similarly, the gene coding for red flower is B and its recessive allele b codes for white flower.

b). Pure breeding cross, AABB* aabb -----> AaBb ---> F1

Test cross of F1 individuals, AaBb*AaBb ---> AaBb (9/16, tall, red), A_ bb (3/16 tall, white), aa B_ (3/16, short, red), aabb (1/16, short, white).

Means all the F2 individuals are in 9:3:3:1 proportion.

The expected frequencies are,

Tall, red -à 2360 *9/16 = 1327.5

Tall, white -à 2360 *3/16 = 442.5

Short, red -à 2360 *3/16 = 442.5

Short, white -à 2360 *1/16 = 147.5

b). CHI - SQUARE (X²):

X² = (O - E)² / E

Where O = Observed frequency

E = Expected frequency

Phenotype

O

E

(O-E)

(O-E)^2

(O-E)^2/E

A_B_

1310

1327.5

-17.5

306.25

0.230697

A_bb

420

442.5

-22.5

506.25

1.144068

aa A_

480

442.5

1.5

2.25

0.155172

aa bb

150

147.5

2.5

6.25

0.042373

1.529937

The calculated Chi-square value is = 1.529

Degrees of freedom is = n-1 = 4-1 = 3

The P value is p < 0.675594, which is not significant at p< 0.05. So, we accept the null hypothesis.

The difference between the expected an observed phenotypic ratios is not significant and the genes are assorting independently

c).

Pure breeding cross, AABB* aabb -----> AaBb ---> F1

Test cross of F1 individuals, AaBb*AaBb ---> AaBb (9/16, tall, red), A_ bb (3/16 tall, white), aa B_ (3/16, short, red), aabb (1/16, short, white).

Phenotype	O	E	(O-E)	(O-E)^2	(O-E)^2/E
A_B_	1310	1327.5	-17.5	306.25	0.230697
A_bb	420	442.5	-22.5	506.25	1.144068
aa A_	480	442.5	1.5	2.25	0.155172
aa bb	150	147.5	2.5	6.25	0.042373
					1.529937