Biology

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

quantities:

(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

(G)

Breeding value

(A)

Dominance deviation

(D)

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,

11

12

22

1 1 0

1 0 1

1 1 0

G

G a

G d

           

                 

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.

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