AABB AABb AaBB AaBb AAbb Aabb aaBB aaBb aa bb Complete domin

AABB

AABb

AaBB

AaBb

AAbb

Aabb

aaBB

aaBb

aa bb

Complete dominance

9

3

3

1

Recessive epistasis

9

3

4

Dominant epistasis

12

3

1

Suppression

13a

3

Complementation

9

7

Duplicate genes

15

1

^aThe 13 is composed of the 12 immediately above plus the one aa bb from the last column.

Solve the following. The following are dihybrid F2 data: A_B_ = 406; A_bb = 149; aaB_ = 162; and aabb = 51. Propose a hypothesis/mode of inheritance for the observed outcome. Test the hypothesis by using the chi-square analysis.

	AABB	AABb	AaBB	AaBb	AAbb	Aabb	aaBB	aaBb	aa bb
Complete dominance		9	3		3	1
Recessive epistasis		9	3		4
Dominant epistasis		12		3	1
Suppression		13a		3
Complementation		9			7
Duplicate genes		15			1

Solution

In the given table [the observed progenies]

A_B_ = 406

A_bb = 149

aaB_ = 162

aabb = 51

Total= 768

According the above observations, we can make hypothesis that the test may not follow rule of independent assortment because of a significant linkage in ‘a’ and ‘b’ alleles.

The hypothesis has made on the basis of observable ratio.

To find the chi square test se the following example:

Chi square (X²) test = (observed- expected)² /expected = (O-E)²/ E

alleles

Observed population

Expected population

(O-E)²

X² values

M

76

50

676

13.52

N

36

50

196

3.92

MN

88

100

144

1.44

Total chi square value

18.88

To find the expected population,

Use Hardy Weinberg law which states two alleles have combined 1 frequency.

P+Q =1

According the rule of independent assortment

(P+Q)² = P² +Q²+ 2PQ in which P and Q would have =0.25 frequency and PQ will share 0.50 frequency.

alleles	Observed population	Expected population	(O-E)2	X2 values
M	76	50	676	13.52
N	36	50	196	3.92
MN	88	100	144	1.44
Total chi square value				18.88