Question

Researchers are interested in how environment and interaction effects influence intelligence. In a colony of laboratory rats that are reared in an enriched environment (e.g. they have a varying number of toys, problem-solving activities, and food options) the variance in maze-solving time amongst individual rats is 5 min2. From this colony, they have also derived three inbred lines for low, medium and high intelligence. When individuals from these different inbred lines are reared together under strict uniform laboratory control conditions (i.e. with no environmental enrichment and standard formulated feed) the variance in maze-solving time is 1.5 min2. When individuals from only one of these inbred lines are reared in the enriched environment the variance is 2.1 min2. With regards to this trait in the rat colony, what percentage of total phenotypic variance could be ascribed to (a) Genetics effects; (b) Environmental effects; and (c) Gene-by-Environmental interaction?

(a) 30%

(b) 42%

(c) 28%

I know the answers form the memo, but they do not explain why they use the values in their calculations. Can you please explain the theory behind the values used?

Answer 1

Now, phenotypic variance (V_p) gives an idea about phenotypic variation in population due to combination of genetic (V_g) and environmental factors (V_e).

Mathematically, Total phenotypic variance (V_p) = genetic effect (V_g) + environmental effect (V_e) + (V_g x V_e)

where (V_g x V_e) indicates the variance due to gene-environmental interaction

Now, Phenotypic variance (given) = 5 min²

This is considered as phenotypic variance because these rats are reared with enriched environment (effect of environment is there on intelligence) and is of a random group (effect of genes would be there).

Variance due to effect of genes only = 1.5 min²

This is taken so because it is mentioned that no environmental enrichment was there

Variance due to effect of environment only = 2.1 min²

This is so because rats were from one inbred line so would have same genes (no variation in genotype) and had enriched environment

Percentage ascribed to genetic effects = (Variance due to effect of genes/phenotypic variance) x 100

Percentage ascribed to genetic effects = (1.5/5.0)x 100 = 30 %

Percentage ascribed to environmental effects = (Variance due to effect of environment/phenotypic variance) x 100

Percentage ascribed to genetic effects = (2.1/5.0)x 100 = 42 %

Now, remaining is due to interaction between gene and environment

(V_g x V_e) = V_p - (V_g +V_e)

(V_g x V_e) = 5.0 - (1.5+2.1)

(V_g x V_e) = 1.4

Percentage ascribed to gene-environment interaction = [(V_g x V_e)/V_p] x 100

Percentage ascribed to gene-environment interaction = [1.4/5.0] x 100 = 28 %