$$g_1=(.6*x_{8}+.5*x_{11}-.4*x_{14})$$
$$if (pc_1<0) \left[ g_2=.6*x_{208}-.5*x_{211}-.4*x_{214}\right]$$
else $$\left[ g_2=-(.6*x_{208}+.5*x_{211}+.4*x_{214})\right]$$
$$g_3=(.6*x_{408}+.5*x_{411}-.4*x_{414})^2$$
$$if(pc_1<0) \left[ g_4=((.6*x_{608}+.5*x_{611}-.4*x_{614})^2)\right]$$
else $$\left[ g_4=(-(.6*x_{608}-.5*x_{611}+.4*x_{614})^2)\right]$$
$$if(pc_1<0) \left[ g_5=((.6*x_{808}+.5*x_{811}-.4*x_{814}+.5*pc_2)^2)\right]$$
else $$\left[ g_5=((-.6*x_{808}-.5*x_{811}-4*x_{814}+.5*pc_2)^2)\right]$$
1. Five genetic effects $$g_i, i=1,\ldots , 5$$ at five loci were generated according to the formulas below. Each effect was standardized to have a variance of 1 over the simulated genotypes and the total genotypic value of an genotype was calculated as the sum of these effects. The individuals were evenly assigned to one of the two sexes at random, which in turn was reflected in the genetic values as a fixed difference of 5 units. The final phenotypes for the individuals were obtained by adding independent and identically distributed, zero centered normal random variables to genetic values to obtain a broad-sense heritability of 2/3