**Practice 5**

The Flemish breed of rabbits has an average body weight of 3600 grams (g). The Himalayan breed has a mean of 1875 g. Matings between these two breeds produce an intermediate F_{1} with a standard deviation of 162 g. The variability of the F_{2} is greater, as indicated by a standard deviation of 230 g. (*a*) Estimate the number of pairs of factors contributing to mature body weight in rabbits. (*b*) Estimate the average quantitative contribution of each active allele.

**Solution 5**

- From formula (8.6),
- The difference 3600 – 1875 = 1725 g is attributed to 14
*pairs*of factors or 28 active alleles. The average contribution of each active allele is

1725/28 = 61:61 g.

**Practice 6**

In a normally distributed population that has a phenotypic mean of 55 units, a total genetic variance for the trait of 35 units^{2}, and an environmental variance of 14 units^{2}, between what two phenotypic values will approximately 68% of the population be found?

**Solution 6**

68% of a normally distributed population is expected to be found within the limits μ ± σ = 55 ± 7, or between 48 units and 62 units.

**Practice 7**

Identical twins are derived from a single fertilized egg (monozygotic). Fraternal twins develop from different fertilized eggs (dizygotic) and are expected to have about half of their genes in common. Left-hand middle-finger-length measurements were taken on the fifth birthday in samples (all of the same sex) of identical twins, fraternal twins, and unrelated individuals from a population of California Caucasians. Using only variances between twins and between unrelated members of the total population, devise a formula for estimating the heritability of left-hand middle-finger length at 5 years of age in this population.

**Solution 7**

Let *V _{i}* = phenotypic variance between identical twins,

*V*= phenotypic variance between fraternal twins, and

_{f}*V*= phenotypic variance between randomly chosen pairs from the total population of which these twins are a part. All of the phenotypic variance between identical twins is nongenetic (environmental); thus,

_{t}*V*=

_{i}*V*. The phenotypic variance between fraternal twins is partly genetic and partly environmental. Since fraternal twins are 50% related, their genetic variance is expected to be only half that of unrelated individuals; thus,

_{e}*V*= (1/2)

_{f}*V*+

_{g}*V*. The difference (

_{e}*V*–

_{f}*V*) estimates half of the genetic variance.

_{i}Therefore, heritability is twice that difference divided by the total phenotypic variance.

**Practice 8**

Two homozygous varieties of *Nicotiana longiflora* were crossed to produce F_{1} hybrids. The average variance of corolla length for all three populations was 8.76. The variance of the F_{2} was 40.96. Estimate the heritability of flower length in the F_{2} population.

**Solution 8**

Since the two parental varieties and the F_{1} are all genetically uniform, their average phenotypic variance is an estimate of the environmental variance (*V _{e}*). The phenotypic variance of the F

_{2}(

*V*) is partly genetic and partly environmental. The difference (

_{t}*V*–

_{t}*V*) is the genetic variance (

_{e}*V*).

_{g}**Practice 9**

The heading rate data are recorded in the table below for two pure varieties of wheat, their F_{1} and F_{2} progenies, and the first backcross generations.

Find: (*a*) the best estimate of the environmental variance (*V _{E}*), (

*b*) the broad heritability (

*H*) for this trait in this population, (

^{2}*c*) the additive genetic variance for this trait, (

*d*) the narrow heritability (

*h*) estimate, (

^{2}*e*) the dominance variance (

*V*), and (

_{D}*f*) the degree of dominance.

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