Review the following concepts if needed:
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.
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
- 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 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 (Vf – Vi) 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 (Vt – Ve) 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.
Solution 9
- Since all of the phenotypic variance within pure lines and their genetically uniform F1 progeny is environmental, the mean of these variances produces the best estimate of the environmental variance (VE), assuming that there has been no change in the environment from one generation to the next.
- The phenotypic variance of the F2 has a genetic and an environmental component.
- The total genetic variance has both an additive and a dominance component. From page 270 we have
where A = sum of all squared additive deviations from the mean, 1/2A = VA and 1/4D = VD. Multiplying formula (1) by a factor of 2 and subtracting formula (2), we have
- The narrow heritability estimate (h2) is
h2 = VA/VF2 = 29:06/40:35 = 0:72
Notice that h2 is smaller than H2, calculated in part (b). This indicates that most of the genetic variance is additive and relatively less is due to dominance variance.
- The dominance variance is calculated thus:
- In step (c) the following formula was given
Thus, if (1/2)A = VA and (1/2)D = VD, then A = 2VA and D = 4VD, and the degree of dominance (formula (8.10)) is
Practice 10
The pounds of grease fleece weight was measured in a sample from a sheep population. The data listed below is for the average of both parents (X, midparent) and their offspring (Y).
(a) Calculate the regression coefficient of offspring on midparent and estimate the heritability of grease fleece weight in this population. (b) Plot the data and draw the regression line. (c) Calculate the correlation coefficient and from that estimate the heritability.
Solution 10
The regression of offspring on midparent is an estimate of heritability:
h2 = 0:67
- Data plot and regression line:
The regression line goes through the intersection of the two means (, 
); = 99:8/10 9:98; = 67:3/10 = 6:73. The regression line intersects the Y axis at the Y intercept (a; formula 8.12).
a = – b = 6:73 – 0:673 (9:98) = 0
Now let us choose a value of X that is distant from (, ) but easily plotted on the graph (e.g., X = 8:0). The corresponding value of Y is estimated to be
= a + bX = 0:01 + 0:673 (8:0) = 5:39
These two points [(, ) and ] establish the regression line with slope b = 0:67. For every 1 lb increase in midparent values, offspring tend to produce 0.67 lb.
- The correlation coefficient (r) is calculated using formula (8.18):
Therefore, the X and Y values are very highly positively correlated. Note that two variables can be highly correlated without also being nearly equal. Two variables are perfectly correlated if for one unit change in one variable there is a constant change (either plus or minus) in the other. Negative correlations for heritability estimates are biologically meaningless. Different traits, however, may have negative genetic correlations (e.g., total milk production vs. butterfat percentage in dairy cattle); many of the same genes that contribute positively to milk yield also contribute negatively to butterfat content.
Practice 11
The total genetic variance of 180-day body weight in a population of swine is 250 lb2. The variance due to dominance effects is 50 lb2. The variance due to epistatic effects is 20 lb2. The environmental variance is 350 lb2. What is the heritability estimate (narrow sense) of this trait?
Solution 11
Practice 12
The heritability of feedlot rate of gain in beef cattle is 0.6. The average rate of gain in the population is 1:7 lb/day. The average rate of gain of the individuals selected from this population to be the parents of the next generation is 2:8 lb/day. What is the expected average daily gain of the progeny in the next generation?
Solution 12
Practice 13
Fifty gilts (female pigs) born each year in a given herd can be used for proving sires. Average litter size at birth is 10 with 10%mortality to maturity. Only the 5 boars (males) with the highest sire index will be saved for further use in the herd. If each test requires 18 mature progeny, how much culling can be practiced among the progeny-tested boars; i.e., what proportion of those tested will not be saved?
Solution 13
Each gilt will produce an average of 10 – (0:1) (10) = 9 progeny raised to maturity. If 18 mature progeny are required to prove a sire, then each boar should be mated to 2 gilts. (50 gilts) / (2 gilts per boar) = 25 boars can be proved. 20/25 = 4/5 = 80% of these boars will be culled.
Practice 14
Given the following pedigree with butterfat records on the cows and equal-parent indices on the bulls, estimate the index for young bull X (a) using information from A and B, (b) when the record made by B is only one lactation, and that made in another herd.
Solution 14
- The midparent index (estimate of transmitting ability) for X is (750 + 604)/2 = 677.
- Since we cannot rely on B's record, we should use information from C and D, recalling that X is separated by two Mendelian segregations from the grandparents. Then X = 750/2 + 820/ 4 + 492/4 = 703.
Practice 15
Calculate the inbreeding coefficient for A in the following pedigree.
Solution 15
First we must convert the pedigree to an arrow diagram.
There is only one pathway from B to C and that goes through ancestor E. However, ancestor E is himself inbred. Note that the parents of E are full sibs; i.e., G and H are 50% related (see Example 8.17). By formula (8.23),
FE = (1/2)RGH = 1/2 (0:5) = 0:25
The inbreeding coefficient of A is given by equation (8.26),
FA = (1/2)n (1 + Fancestor) = (1/2)3 (1 + 0:25) = 0:156
where n is the number of arrows connecting the individual (A) through one parent (B) back to the common ancestor (E) and back again to the other parent (C).
Practice 16
The average plant heights of two inbred tobacco varieties and their hybrids have been measured with the following results: inbred parent (P1 ) = 47:8 in, inbred parent (P2) = 28:7 in, F1 hybrid (P1 × P2) = 43:2 in. ´ (a) Calculate the amount of heterosis exhibited by the F1. (b) Predict the average height of the F2.
Solution 16
- The amount of heterosis (formula (8.27)) is expressed by the excess of the F1 average over the midpoint between the two parental means.
- As a general rule, the F2 shows only about half the heterosis of the F1: 1/2 (4:95) = 2:48. Hence, the expected height of F2 plants = 38:25 + 2:48 = 40:73 in.
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