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Chapter 23
23.1 If heart disease is considered to be a threshold trait, what genetic and
environmental factors might contribute to the underlying liability for a person to develop
this disease?
ANS: Some of the genes implicated in heart disease are listed in Table 23.2.
Environmental factors might include diet, amount of exercise, and whether or not the
person smokes.
FEEDBACK: 23.1
DIFFICULTY: easy
23.2. A wheat variety with red kernels (genotype AA BB) was crossed with a variety
with white kernels (genotype AA BB). The F1 were intercrossed to produce an F2. If each
primed allele increases the amount of pigment in the kernel by an equal amount, what
phenotypes will be expected in the F2? Assuming that the A and B loci assort
independently, what will the phenotypic frequencies be?
ANS: 1/16 red; 4/16=1/4 dark pink; 6/16 pink; 4/16=1/4 light pink; 1/16 white
FEEDBACK: 23.1
DIFFICULTY: medium
23.3 For alcoholism, the concordance rate for monozygotic twins is 55 percent,
whereas for dizygotic twins, it is 28 percent. Do these data suggest that alcoholism has a
genetic basis?
ANS: The concordance for monozygotic twins is almost twice as great as that for
dizygotic twins. Monozygotic twins share twice as many genes as dizygotic twins. The
data strongly suggest that alcoholism has a genetic basis.
FEEDBACK: 23.1
DIFFICULTY: easy
23.4. The height of the seed head in wheat at maturity is determined by several genes.
In one variety, the head is just 9 inches above the ground; in another, it is 33 inches above
the ground. Plants from the 9-inch variety were crossed to plants from the 33-inch
variety. Among the F1, the seed head was 21 inches above the ground. After selffertilization, the F1 plants produced an F2 population in which 9-inch and 33-inch plants
each appeared with a frequency of 1/256. (a) How many genes are involved in the
determination of seed head height in these strains of wheat? (b) How much does each
allele of these genes contribute to seed head height? (c) If a 21-inch F1 plant were crossed
to a 9-inch plant, how often would you expect 18-inch wheat to occur in the progeny?
ANS: (a) 4; (b) 3 inches; (c) frequency of 1/4.
FEEDBACK: 23.2
DIFFICULTY: hard
23.5 Assume that size in rabbits is determined by genes with equal and additive effects.
From a total of 2012 F2 progeny from crosses between true-breeding large and small
varieties, eight rabbits were as small as the small variety and eight were as large as the
large variety. How many size-determining genes were segregating in these crosses?
ANS: Because 8/2012 is approximately 1/256 = (1/4)4, it appears that four sizedetermining genes were segregating in the crosses.
FEEDBACK: 23.2
DIFFICULTY: medium
23.6. A sample of 20 plants from a population was measured in inches as follows: 18,
21, 20, 23, 20, 21, 20, 22, 19, 20, 17, 21, 20, 22, 20, 21, 20, 22, 19, and 23. Calculate (a)
the mean, (b) the variance, and (c) the standard deviation.
ANS: (a) The mean is 20.45 inches. (b) The variance is 2.37 inches2. (c) The standard
deviation is 1.54 inches.
FEEDBACK: 23.2
DIFFICULTY: easy
23.7 Quantitative geneticists use the variance as a measure of scatter in a sample of
data; they calculate this statistic by averaging the squared deviations between each
measurement and the sample mean. Why don’t they simply measure the scatter by
computing the average of the deviations without bothering to square them?
ANS: Because (Xi  mean) = 0
FEEDBACK: 23.2
DIFFICULTY: medium
23.8. Two inbred strains of corn were crossed to produce an F1, which was then
intercrossed to produce an F2. Data on ear length from a sample of F1 and F2 individuals
gave phenotypic variances of 15.2 cm2 and 27.6 cm2, respectively. Why was the
phenotypic variance greater for the F2 than for the F1?
ANS: For the F1, Vg = 0 because they are all genetically identical and heterozygous; for
the F2, Vg > 0 because genetic differences result from the segregation and independent
assortment of genes.
FEEDBACK: 23.3
DIFFICULTY: medium
23.9 A study of quantitative variation for abdominal bristle number in female
Drosophila yielded estimates of VT = 6.08, Vg = 3.17, and Ve = 2.91. What was the broadsense heritability?
ANS: 3.17/6.08 = 0.52
FEEDBACK: 23.3
DIFFICULTY: easy
23.10. A researcher has been studying kernel number on ears of corn. In one highly
inbred strain, the variance for kernel number is 426. Within this strain, what is the broadsense heritability for kernel number?
ANS: The broad –sense heritability within a highly inbred strain is expected to be zero
because there is no genetic variability.
FEEDBACK: 23.3
DIFFICULTY: medium
23.11 Measurements on ear length were obtained from three populations of corn—two
inbred varieties and a randomly pollinated population derived from a cross between the
two inbred strains. The phenotypic variances were 9.2 cm2 and 9.6 cm2 for the two inbred
varieties and 26.4 cm2 for the randomly pollinated population. Estimate the broad-sense
heritability of ear length for these populations.
ANS: Ve is estimated by the average of the variances of the inbreds: 9.4 cm 2. Vg is
estimated by the difference between the variances of the randomly pollinated population
and the inbreds: (26.4  9.4) = 7.4 cm2. The broad-sense heritability is H2 = Vg/Vt =
7.4/26.4 = 0.64.
FEEDBACK: 23.2
DIFFICULTY: medium
23.12. Figure 23.4 summarizes data on maturation time in populations of wheat. Do
these data provide any insight as to whether or not this trait is influenced by dominance?
Explain.
ANS: Because the F1 plants have maturation times midway between those of the parental
strains, there seems to be little or no dominance for this trait.
FEEDBACK: 23. 3 and 23.4
DIFFICULTY: medium
23.13 A quantitative geneticist claims that the narrow-sense heritability for body mass
in human beings is 0.7, while the broad-sense heritability is only 0.3. Why must there be
an error?
ANS: Broad-sense heritability must be greater than narrow-sense heritability because H2
= Vg/Vt  Va/Vt = b2.
FEEDBACK: 23.3
DIFFICULTY: medium
23.14. The mean value of a trait is 100 units, and the narrow-sense heritability is 0.4. A
male and a female measuring 124 and 126 units, respectively, mate and produce a large
number of offspring, which are reared in an average environment. What is the expected
value of the trait among these offspring?
ANS: (125 – 100)(0.4) + 100 = 110 units.
FEEDBACK: 23.3
DIFFICULTY: easy
23.15 The narrow-sense heritability for abdominal bristle number in a population of
Drosophila is 0.3. The mean bristle number is 12. A male with 10 bristles is mated to a
female with 20 bristles, and a large number of progeny are scored for bristle number.
What is the expected number of bristles among these progeny?
ANS: (15  12)(0.3) + 12 = 12.9 bristles.
FEEDBACK: 23.3
DIFFICULTY: medium
23.16. A breeder is trying to decrease the maturation time in a population of sunflowers.
In this population, the mean time to flowering is 100 days. Plants with a mean flowering
time of only 90 days were used to produce the next generation. If the narrow-sense
heritability for flowering time is 0.2, what will the average time to flowering be in the
next generation?
ANS: (90 – 100)(0.2) + 100 = 98 days
FEEDBACK: 23.3
DIFFICULTY: medium
25.17 A fish breeder wishes to increase the rate of growth in a stock by selecting for
increased length at six weeks after hatching. The mean length of six-week-old fingerlings
is currently 10 cm. Adult fish that had a mean length of 15 cm at six weeks of age were
used to produce a new generation of fingerlings. Among these, the mean length was 12.5
cm. Estimate the narrow-sense heritability of fingerling length at six weeks of age and
advise the breeder about the feasibility of the plan to increase growth rate.
ANS: b2 = R/S = (12.5  10)/(15  10) = 0.5; selection for increased growth rate should
be effective.
FEEDBACK: 23.3
DIFFICULTY: medium
23.18. Leo’s IQ is 86 and Julie’s IQ is 110. The mean IQ in the population is 100.
Assume that the narrow-sense heritability for IQ is 0.4. What is the expected IQ of Leo
and Julie’s first child?
ANS: (98 – 100)(0.4) + 100 = 99.2
FEEDBACK: 23.4
DIFFICULTY: easy
23.19 One way to estimate a maximum value for the narrow-sense heritability is to
calculate the correlation between half-siblings that have been reared apart and divide it by
the fraction of genes that half-siblings share by virtue of common ancestry. A study of
human half-siblings found that the correlation coefficient for height was 0.14. From this
result, what is the maximum value of the narrow-sense heritability for height in this
population?
ANS: Half-siblings share 25 percent of their genes. The maximum value for h2 is
therefore 0.14/0.25 = 0.56.
FEEDBACK: 23.4
DIFFICULTY: medium
23.20. A selection differential of 40 g per generation was used in an experiment to
select for increased pupa weight in Tribolium. The narrow-sense heritability for pupa
weight was estimated to be 0.3. If the mean pupa weight was initially 2000 g and
selection was practiced for 10 generations, what was the mean pupa weight expected to
become?
ANS: F1 = (1/2)3 (1 + FC) = 0.25; thus, FC = 1.
FEEDBACK: 23.3
DIFFICULTY: medium
23.21 On the basis of the observed correlations for personality traits shown in Table
23.5, what can you say about the value of the environmentality (C2 in Table 23.3)?
ANS: The correlations for MZT are not much different from those for MZA. Evidently,
for these personality traits, the environmentality (C2 in Table 23.3) is negligible.
FEEDBACK: 23.5
DIFFICULTY: medium