Quantitative Genetics Practice Problems (page 2)
Review the following concepts if needed:
- Qualitative vs. Quantitative Traits for Genetics
- Polygenic Traits for Genetics
- The Normal Distribution for Genetics
- Types of Gene Action for Genetics
- Heritability for Genetics
- Selection Methods for Genetics
- Mating Methods for Genetics
Quantitative Genetics Practice Problems
Two homozygous varieties of Nicotiana longiflora have mean corolla lengths of 40.5 and 93.3 mm. The average of the F1 hybrids from these two varieties was of intermediate length. Among 444 F2 plants, none was found to have flowers either as long as or as short as the average of the parental varieties. Estimate the minimal number of pairs of alleles segregating from the F1.
If four pairs of alleles were segregating from the F1, we expect (1/4)4 = 1/256 of the F2 to be as extreme as one or the other parental average. Likewise, if five pairs of alleles were segregating, we expect (1/4)5 = 1/1024 of the F2 to be as extreme as one parent or the other. Since none of the 444 F2 plants had flowers this extreme, more than four loci (minimum of five loci) are probably segregating from the F1.
The mean internode length of the Abed Binder variety of barley was found to be 3.20mm. The mean length in the Asplund variety was 2.10 mm. Crossing these two varieties produced an F1 and F2 with average internode lengths of 2.65 mm. About 6%of the F2 had an internode length of 3.2 mm, and another 6% had a length of 2.1 mm. Determine the most probable number of gene pairs involved in internode length and the approximate contribution each gene makes to the phenotype.
With one pair of genes we expect about 1/4 or 25% of the F2 to be as extreme as one of the parents. With two pairs of genes we expect approximately 1/16 or 6.25% as extreme as one parent. Thus, we may postulate two pairs of genes. Let A and B represent growth factors and a and b represent null genes
The difference 2:65 – 2.10 = 0.55mm is the result of two growth genes. Therefore, each growth gene contributes 0.275 mm to the phenotype.
A large breed of chicken, the Golden Hamburg, was crossed to small Sebright Bantams. The F1 was intermediate in size. The mean size of the F2 was about the same as that of the F1, but the variability of the F2 was so great that a few individuals were found to exceed the size of either parental type (transgressive variation). If all of the alleles contributing to size act with equal and cumulative effects, and if the parents are considered to be homozygous, how can these results be explained?
Let capital letters stand for growth genes (active alleles) and small letters stand for alleles that do not contribute to growth (null alleles). For simplicity, we will consider only four loci.
A representative sample of lamb weaning weights is shown below. Determine the weight limits within which 95% of all lambs from this population are expected to be found at weaning time.
The standard deviation is calculated as follows:
95% of all weaning weights are expected to lie within ± 2s of the mean. Thus, ± 2s = 81:0 ± 2(13:77) = 81:0 ± 27:54. The upper limit is 108.54 and the lower limit is 53.46.
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.
- 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.
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?
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.
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.
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 (Vf – Vi) estimates half of the genetic variance.
Therefore, heritability is twice that difference divided by the total phenotypic variance.
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.
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 (Vt – Ve) is the genetic variance (Vg).
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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