Consider a hypothetical quantitative trait ( a weight of some kind) affected by

Question

Consider a hypothetical quantitative trait ( a weight of some kind) affected by seven loci. Assume the following:

Complete dominance at all loci. No epistasis. (3 pts)

The independent effect of each dominant gene is +6 lb.

The independent effect of each recessive gene is -3 lb.

For homozygous combinations, genotypic values are equal to breeding values.

? = 700 lb.

Fill in the following table:

Genotype

BV

G

GCV

E

P

BBeeAaVVEerrSS

+13

BbEeAaVvEeRrSs

-18

Which individual is the heaviest? Explain.

Which would produce the heaviest offspring (on average)? Explain.

BV = breeding values, GCV = gene combination values, E = Environmental effects P = phenotypic values

Genotype

BV

G

GCV

E

P

BBeeAaVVEerrSS

+13

BbEeAaVvEeRrSs

-18

Explanation / Answer

Quantitative trait locus (QTL) that corelates to the variation in phenotypes or quantitative traits. Phenotypic variation may result due to genetic variation, effect of the environment in which the organism exists, or interaction of environment with genetic components.

Thus, to determine, how much phenotypic variation is due to genetic interactions or allelic variance, and how much is due to environmental factor, quantification is required (quantitative genetics or quantitative trait locus or QTL mapping).

Breeding value (BV)- measures the effect of genetic makeup on the phenotype. It is calculated as sum of average effect of alleles.

Thus, breeding value is expressed as the average effect (effect on a gene allele measured in terms of locus and generations).

Gene combination value (GCV), is part of an individual's genotypic value, that is due to effect of gene combination. GCV is related to BV and G as:

G = GCV + BV

Phenotype value is designated as P (describing a trait or phenotype), genetic value or genotype by G, and E is environmental deviation (representing effect of environment and interaction of genotype with environment):

P = G + E

For gene combination values:

P = µ + BV + GCV + E (mean population, 700 lbs in this case).

8(+6)+6(-3)

=48-18=30

4(+12)+3(-6)

=48-18=30

700+30 +0+13

7(+6) + 7 (-3)

= 42 -21

=21

700+21+63 -18

Genotype 2, is heavier, as phenotype or performance value is higher.

Genotype 1 will produce heavier offsprings, as breeding value is higher.

.

Genotype BV G GCV E P BBeeAaVVEerrSS

8(+6)+6(-3)

=48-18=30

4(+12)+3(-6)

=48-18=30

30-30=0 +13

700+30 +0+13

= 743lb BbEeAaVvEeRrSs

7(+6) + 7 (-3)

= 42 -21

=21

7(+12) = 84 84-21=63 -18

700+21+63 -18

=766lb

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