Question

12. Two pure breeding large-grained strains of rye were crossed. All the F1 progeny were small- grained. F2 progeny produced by selfing the F1 small-grained plants segregated 91 small and 69 large.

a. Propose a genetic model to explain these observations, including defining genes, alleles, and allelic relationships.

b. Use the chi-square ( ?2 ) analysis to test your hypothesis (see probability table). Include all calculations, p-value, and an interpretation of your results for full credit.

please explain how you got the numbers

Answer 1

a).

91/69 = 1.31

9/7 = 1.28

The numbers of phenotype indicates that they show 9:7 phenotypic ratio. This ratio is found in duplicate recessive epistasis.

A_B_ = Small

A_bb or aaB_ or aabb = Large

AAbb (large) x (large) aaBB ---Parents

AaBb (small)---------------------F1

AaBb x AaBb --F1 x F1

A_B_ = 9/16

A_bb + aaB_ + aabb = 7/16

Total progeny = 91+69 =160

Expected small = 9/16* 160 = 90

Expected large = 7/16 *160 = 70

b).

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
small	91	90	1	1.0000	0.0111
large	69	70	-1	1.0000	0.0143
Total	160	160			0.0254

X^2 = 0.025

Degrees of freedom = no. of phenotypes - 1

df = 2-1=1

p vlaue = 3.84

The X^2 value of 0.025 is less than the p value of 3.84. Hence, null hypothesis is accepted and the data is significant to the expected values.