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# Quantitative Genetics Practice Problems (page 2)

based on 4 ratings
By — McGraw-Hill Professional
Updated on Aug 23, 2011

#### 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 F1 with a standard deviation of 162 g. The variability of the F2 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

1. From formula (8.6),
2. 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
3. 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 units2, and an environmental variance of 14 units2, 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 Vi = phenotypic variance between identical twins, Vf = phenotypic variance between fraternal twins, and Vt = 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, Vi = Ve. 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, Vf = (1/2) Vg + Ve. The difference (VfVi) estimates half of the genetic variance.

Therefore, heritability is twice that difference divided by the total phenotypic variance.

#### Practice 8

Two homozygous varieties of Nicotiana longiflora were crossed to produce F1 hybrids. The average variance of corolla length for all three populations was 8.76. The variance of the F2 was 40.96. Estimate the heritability of flower length in the F2 population.

#### Solution 8

Since the two parental varieties and the F1 are all genetically uniform, their average phenotypic variance is an estimate of the environmental variance (Ve). The phenotypic variance of the F2 (Vt) is partly genetic and partly environmental. The difference (VtVe) is the genetic variance (Vg).

#### Practice 9

The heading rate data are recorded in the table below for two pure varieties of wheat, their F1 and F2 progenies, and the first backcross generations.

Find:   (a) the best estimate of the environmental variance (VE),   (b) the broad heritability (H2) for this trait in this population,   (c) the additive genetic variance for this trait,   (d) the narrow heritability (h2) estimate,   (e) the dominance variance (VD), and   (f) the degree of dominance.

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