AABB AABb AaBB AaBb AAbb Aabb aaBB aaBb aa bb Complete dominance 9 3 3 1 Recessi
ID: 317346 • Letter: A
Question
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
Explanation / Answer
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