Quantitative Genetics (BPSC148)
Assignment 2 - solutions
Question 1:
Genotype ac ac Ac ac Ac Ac
Genotypic Value 0.010 58.747 84.626
Frequency 0.64 0.32 0.04
(1) ,
0.04 0.5 0.32 0.20p=+×= 1 1 0.20 0.80qp
=
−=− =
(2) 1
2
mid-point (84.626 0.010) 42.318=× + = , 84.626 42.318 42.308a
=
−=,
58.747 42.318 16.429d=−=
, actual Mean is
( ) 2 20.1275M a p q pqd=−+ =− 20.1275 42.318 22.1905
−
+=
(3) 1
: [ ( )] 0.8 [42.308 16.429 (0.8 0.2)] 41.732Ac q a d q p
α
=+−=× + ×− =
2
: [ ( )] 0.2 [42.308 16.429 (0.8 0.2)] 10.433ac p a d q p
α
=− + − =− × + × − =−
12
41.732 ( 10.433) 52.165
α
αα
=−= −− = .
Genotype Frequency A - value D -deviation G - value
Ac Ac 0.04 83.465 -21.029 42.308
Ac ac 0.32 31.299 5.257 16.429
ac ac 0.64 -20.866 -1.314 -42.308
(4)
22
2 2 0.2 0.8 52.165 870.793
A
Vpq
α
==××× =
(5)
22
(2 ) (2 0.2 0.8 16.429) 27.639
D
V pqd==×××=
(6) 870.793 27.639 898.432
GAD
VVV=+= + =
