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# Quantitative Genetics Practice Problems

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

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## Quantitative Genetics Practice Problems

#### Practice 1

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.

#### Solution 1

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.

#### Practice 2

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.

#### Solution 2

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.

#### Practice 3

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?

#### Solution 3

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.

#### Practice 4

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.

#### Solution 4

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.

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