Quantitative Genetics (BIOL148/BPSC148)
Assignment 3 – due Thursday, January 28, 2016
Problem 1. The elements of a 4 8 matrix is given in the following table. Let us denote
the matrix by A.
A col1 col2 col3 col4 col5 col6 col7 col8
row1 -0.6758 1.187895 0.05993 -1.41042 0.249359 0.684817 -1.48278 0.905952
row2 -0.30418 1.308264 -1.10504 0.26642 1.651046 -0.56442 1.910225 1.523252
row3 -0.71968 1.637417 1.079418 -0.09462 0.699059 0.434304 -0.46659 1.606218
row4 -1.18086 -0.69885 -2.15808 1.189356 -0.55777 0.008491 0.759354 1.285645
(1) Calculate A + 0.5A, which is equivalent to (1+0.5)A = 1.5A
(2) Find TA
(3) Calculate TAA
(4) Calculate the determinant of TAA
(5) Calculate T 1(AA )
(6) Calculate ATA
Problem 2. Below is the genetic composition at the acid locus of a citrus population
(Fang et al. 1997). The actual value of each genotype and the genotypic counts are listed
below. Note that the “High” allele is Ac not ac, although ac has a higher allelic frequency.
In the standard notation, Ac is A1 and ac is A2.
Genotype ac ac (A2A2) Ac ac (A2A1) Ac Ac (A1A1)
Genotype value (scale 2) 0.010 58.747 84.626
Genotype counts 640 320 40
Assuming that the population is in Hardy-Weinberg equilibrium, calculate the following
(1) Gene frequencies
(2) Mid-point value ( )
(3) Additive effect (a)
(4) Dominance effect (d)
(5) Population mean (M)
(6) Actual population mean ( M )
(7) Average effects of alleles Ac (A1) and ac (A2)
(8) Average effect of gene substitution
Problem 3. For the same citrus data in Problem 2, the frequency of allele Ac (A1) is
0.2p and thus the frequency of allele ac (A2) is 1 0.2 0.8q . Assuming that the
population is in Hardy-Weinberg equilibrium, you have already calculated the additive
effect 42.308a , the dominance effect 16.426d and the average effect of gene
substitution ( ) 52.165a d q p . The following table gives formulas for the
frequency, the genotypic value, the breeding value and the dominance deviation for each
of the three genotypes.
Table 3. Distribution table of three genotypes of a biallelic population.
Genotype Frequency Genotypic value
1 1A A 2p a 2q 22q d
1 2A A 2 pq d ( )q p 2pqd
2 2A A 2q a 2p 22p d
(1) Replace the expressions in the above table by the actual values. For example, the
frequency of 1 1A A is 2p , which should be replaced by 20.2 0.04 .
(2) Calculate the additive genetic variance ( AV )
(3) Calculate the dominance variance ( DV )
(4) Calculate the total genetic variance ( G A DV V V )
Problem 4. For the calculated µ, a and d from the previous problem (scale 1 in Problem
2), convert the three values (scale 1) back into the three genotypic values (scale 2) using
the following equation,
1 1 0
1 0 1
1 1 0
where 11G is the genotypic value for genotype Ac Ac, 12G is the genotypic value for
genotype Ac ac, and 22G is the genotypic value for genotype ac ac. If your converted
genotypic values are different from the values given in the table of Problem 2, try it again
until they are the same.