Question

Two true breeding strains of peas, one with yellow and round seeds and the other with green and wrinkled seeds, were crossed. All the F1plants had yellow and round seeds. When these F1 plants were test crossed to the green and wrinkled parental strain, the following offspring were obtained: 30 yellow and round; 20 green and round; 26 yellow and wrinkled; 28 green and wrinkled.

A) Please diagram the crosses described above, showing genotypes for each plant, using your own CLEARLY DEFINED nomenclature. Include the genotypes and proportions of each type of plant that you would expect in the F2 generation if the alleles that determine seed color assort independently from the alleles that determine seed shape.

B) Use Chi square analysis to test the hypothesis that the alleles determining seed color assort independently from the alleles determining seed shape Please show your

work along with the items listed below. (This is the part where I get confuse)

VALUE OF X^2

Degrees of freedom and conclusion regarding the hypothesis

Answer 1

(a). Considering two factors, seed color, and seed shape. The Y (yellow) dominates y (green) to decide seed color, and the R allele for round dominates the r allele for the determination of wrinkled seed. The P (Parental) cross is between true-breeding alleles of round green peas and wrinkled yellow peas. The F1 offspring are all round and yellow. In the F2 generation, the genes at the two loci will separate independently. Alternatively, the phenotypic ratio expected for either character is 3:1.

In the F2 generation, 9:3:3:1 is the phenotypic ratio.

(b). Chi-Square Formula

Χ2 = Ʃ (observed value – expected value)2 / (expected value)

Degrees of freedom (df) = n-1 where n is the number of classes

OBSERVED VALUES	EXPECTED VALUES
ROUND YELLOW SEEDS (30)	(9/16) X 30 = 16.875
ROUND GREEN SEEDS (20)	(3/16) X 20 = 3.75
WRINKLED YELLOW SEEDS (26)	(3/16) X 26 = 4.875
WRINKLED GREEN SEEDS (28)	(1/16) X 28 = 1.75
TOTAL SEEDS = 104	TOTAL = 27.25

= (30-16.875)2/ 16.875 + (20- 3.75)2/3.75 + (26-4.875)2/4.875 + (28-1.75)2/1.75

= 10 + 7 + 9 + 3.9

= 29.9 = 30

Number of classes (n) = 4

Degree of freedom = n-1 + 4-1

Chi square value = 32

The probability of the chi square is greater than 0.05. If the value is less than the 0.05, the hypothesis can be accepted and if the value is greater than the 0.05, the hypothesis will be rejected.