CHAPTER 26
Application and Experimental Questions
E1.
Here are data for height and weight among 10 male college students.
Height (cm) Weight (kg)
159
48
162
50
161
52
175
60
174
64
198
81
172
58
180
74
161
50
173
54
A.
Calculate the correlation coefficients for this group.
B.
Is the correlation coefficient statistically significant? Explain.
Answer:
A.
We first need to calculate the standard deviations for height and weight,
and then the covariance for both traits:
Height: variance = 140.02; standard deviation = 11.8
Weight: variance = 121.43; standard deviation = 11.02
Covariance = 123.3
r( X ,Y ) 
CoV( X ,Y )
SDX SDY
r(X,Y) = 123.3
(11.8)(11.02)
r(X,Y) = 0.948
B.
With 8 degrees of freedom, this value is statistically significant. This
means the association between these two variables occurs more frequently than would be
expected by random sampling error. It does not necessarily imply cause and effect.
E2.
The abdomen length (in millimeters) was measured in 15 male Drosophila and
the following data were obtained: 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1,
1.6, 2.3, and 2.2. Calculate the mean, standard deviation, and variance for this population
of male fruit flies.
Answer: To calculate the mean, we add the values together and divide by the total
number.
1.9  2(2.4)  2.(2.1)  3(2.0)  2(2.2)  1.7  1.8  2(2.3)  1.6
15
Mean  2.1
Mean 
The variance is the sum of the squared deviations from the mean divided by N – 1. The
mean value of 2.1 must be subtracted from each value, and then the square is taken.
These 15 values are added together and then divided by 14 (which is N – 1).
0.85
14
 0.061
Variance=
The standard deviation is the square root of the variance.
Standard deviation = 0.25
E3.
In one strain of cabbage, you conduct an RFLP analysis with regard to head
weight; you determine that seven QTLs affect this trait. In another strain of cabbage, you
find that only four QTLs affect this trait. Note that both strains of cabbage are from the
same species, although they may have been subjected to different degrees of inbreeding.
Explain how one strain can have seven QTLs and another strain four QTLs for exactly
the same trait. Is the second strain missing three genes?
Answer: The heritability reflects the amount of genetic variation that influences a trait. In
the strain with seven QTLs, there are at least seven different genes that exist in two or
more alleles that are influencing the outcome of the trait. In the other strain, there are the
same types of genes, but three of them are monomorphic and therefore do not contribute
to the variation in the outcome of the trait.
E4.
From an experimental viewpoint, what does it mean to say that an RFLP is
associated with a trait? Let’s suppose that two strains of pea plants differ in two RFLPs
that are linked to two genes governing pea size. RFLP-1 is found in 2,000 bp and 2,700
bp bands, and RFLP-2 is found in 3,000 bp and 4,000 bp bands. The plants producing
large peas have RFLP-1 (2,000 bp) and RFLP-2 (3,000 bp); those producing small peas
have RFLP-1 (2,700 bp) and RFLP-2 (4,000 bp). A cross is made between these two
strains, and the F1 offspring are allowed to self-fertilize. Five phenotypic classes are
observed: small peas, small-medium peas, medium peas, medium-large peas, and large
peas. We assume that each of the two genes makes an equal contribution to pea size and
that the genetic variance is additive. Draw a gel and explain what RFLP banding patterns
you would expect to observe for these five phenotypic categories. Note: Certain
phenotypic categories may have more than one possible banding pattern.
Answer: When we say that an RFLP is associated with a trait, we mean that a gene that
influences a trait is closely linked to an RFLP. At the chromosomal level, the gene of
interest is so closely linked to the RFLP that a crossover almost never occurs between
them.
Note: Each plant inherits four RFLPs, but it may be homozygous for one or two of them.
Small: 2,700 and 4,000 (homozygous for both)
Small-medium: 2,700 (homozygous), 3,000, and 4,000; or 2,000, 2,700, and 4,000
(homozygous)
Medium: 2,000 and 4,000 (homozygous for both); or 2,700 and 3,000 (homozygous for
both); or 2,000, 2,700, 3,000, and 4,000
Medium-large: 2,000 (homozygous), 3,000, and 4,000; or 2,000, 2,700, and 3,000
(homozygous)
Large: 2,000 and 3,000 (homozygous for both)
E5.
Let’s suppose that two strains of pigs differ in 500 RFLPs. One strain is much
larger than the other. The pigs are crossed to each other, and the members of the F1
generation are also crossed among themselves to produce an F2 generation. Three distinct
RFLPs are associated with F2 pigs that are larger. How would you interpret these results?
Answer: These results suggest there are (at least) three different genes that influence the
size of pigs. This is a minimum estimate because a QTL may have two or more closely
linked genes. Also, it is possible that the large and small strains have the same RFLP
band that is associated with one or more of the genes that affect size.
E6.
Outline the steps you would follow to determine the number of genes that
influence the yield of rice. Describe the results you might get if rice yield is governed by
six different genes.
Answer: Let’s assume there is an extensive molecular marker map for the rice genome.
We would begin with two strains of rice, one with a high yield and one with a low yield,
that greatly differ with regard to the molecular markers that they carry. We would make a
cross between these two strains to get F1 hybrids. We would then backcross the F1
hybrids to either of the parental strains and then examine hundreds of offspring with
regard to their rice yields and molecular markers. In this case, our expected results would
be that six different markers in the high-producing strain would be correlated with
offspring that produce higher yields. We might get fewer than six bands if some of these
genes are closely linked and associate with the same marker. We also might get fewer
than six if the two parental strains have the same marker that is associated with one or
more of the genes that affect yield.
E7.
In a wild strain of tomato plants, the phenotypic variance for tomato weight is 3.2
g2. In another strain of highly inbred tomatoes raised under the same environmental
conditions, the phenotypic variance is 2.2 g2. With regard to the wild strain:
A.
Estimate VG.
B.
What is hB2?
C.
Assuming all the genetic variance is additive, what is hN2 ?
Answer:
A.
B.
C.
If we assume that the highly inbred strain has no genetic variance:
VG (for the wild strain) = 3.2 g2 – 2.2 g2 = 1.0 g2
HB2 = 1.0 g2/3.2 g2 = 0.31
It is the same as HB2 so it also equals 0.31.
E8.
The average thorax length in a Drosophila population is 1.01 mm. You want to
practice selective breeding to make larger Drosophila. To do so, you choose 10 parents
(five males and five females) of the following sizes: 0.97, 0.99, 1.05, 1.06, 1.03, 1.21,
1.22, 1.17, 1.19, and 1.20. You mate them and then analyze the thorax sizes of 30
offspring (half male and half female).
0.99, 1.15, 1.20, 1.33, 1.07, 1.11, 1.21, 0.94, 1.07, 1.11, 1.20, 1.01, 1.02, 1.05, 1.21, 1.22,
1.03, 0.99, 1.20, 1.10, 0.91, 0.94, 1.13, 1.14, 1.20, 0.89, 1.10, 1.04, 1.01, 1.26
Calculate the realized heritability in this group of flies.
Answer:
R
S
R  XO  X
hN2 
S  XP  X
In this problem:
X equals 1.01 (as given in the problem)
X O equals 1.11 (by calculating the mean for the parents)
X P equals 1.09 (by calculating the mean for the offspring)
R = 1.09 – 1.01 = 0.08
S = 1.11 – 1.01 = 0.10
hN2 = 0.08/0.10 = 0.8 (which is a pretty high heritability value)
E9.
In a strain of mice, the average 6-week body weight is 25 g and the narrow-sense
heritability for this trait is 0.21.
A.
What would be the average weight of the offspring if parents with a mean
weight of 27 g were chosen?
B.
What weight of parents would you have to choose to obtain offspring with
an average weight of 26.5 g?
Answer:
A.
hN 2 
Xo  X
XP  X
0.21  (26.5 g - 25 g)/(27 g - 25 g)
X O  25 g = 2 g (0.21)
X O  25.42 g
B.
0.21  (26.5 g  25 g)/( X p  25 g)
( X p  25 g)(0.21)=1.5 g
X p  32.14 g parents
However, because this value is so far from the mean, there may not be 32.14 g parents in
the population of mice that you have available.
E10. You will need to understand Solved problem S4 before answering this question.
The variance for fathers (in square inches) was 112, the variance for sons was 122, and
the covariance was 144. The mean height for fathers was 68 in., and the mean height for
sons was 69 in. If a father had a height of 70 in., what is the most probable height of his
son?
Answer: We first need to calculate a and b. In this calculation, X represents the height of
fathers and Y represents the height of sons.
144
 1.29
112
a  69  (1.29)(68)  18.7
b
For a father who is 70 in tall,
Y = (1.29)(70) + (–18.7) = 71.6
The most likely height of the son would be 71.6 in.
E11. A danger in computing heritability values from studies involving genetically
related individuals is the possibility that these individuals share more similar
environments than do unrelated individuals. In the experiment of Figure 26.8, which data
are the most compelling evidence that ridge count is not caused by genetically related
individuals sharing common environments? Explain.
Answer: The identical and fraternal twins, which probably share very similar
environments, but who differ in the amount of genetic material they share, are a strong
argument against an environmental bias. The differences in the observed correlations
(0.49 versus 0.99) are consistent with the differences in the expected correlations (0.5
versus 1.0).
E12. A large, genetically heterogeneous group of tomato plants was used as the
original breeding stock by two different breeders, named Mary and Hector. Each breeder
was given 50 seeds and began an artificial selection strategy, much like the one described
in Figure 26.11. The seeds were planted, and the breeders selected the 10 plants with the
highest mean tomato weights as the breeding stock for the next generation. This process
was repeated over the course of 12 growing seasons, and the following data were
obtained:
Mean Weight of Tomatoes (lb)
Year Mary’s Tomatoes Hector’s Tomatoes
1
0.7
0.8
2
0.9
0.9
3
1.1
1.2
4
1.2
1.3
5
1.3
1.3
6
1.4
1.4
7
1.4
1.5
8
1.5
1.5
9
1.5
1.5
10
1.5
1.5
11
1.5
1.5
12
1.5
1.5
A.
Explain these results.
B.
Another tomato breeder, named Martin, got some seeds from Mary’s and
Hector’s tomato strains (after 12 generations), grew the plants, and then crossed them to
each other. The mean weight of the tomatoes in these hybrids was about 1.7 lb. For a
period of 5 years, Martin subjected these hybrids to the same experimental strategy that
Mary and Hector had followed, and he obtained the following results:
Mean Weight of tomatoes (lb)
Year Martin’s Tomatoes
1
1.7
2
1.8
3
1.9
4
2.0
5
2.0
Explain Martin’s data. Is heterosis occurring? Why was Martin able to obtain tomatoes
that are heavier than 1.5 lb, while Mary’s and Hector’s strains appeared to plateau at this
weight?
Answer:
A.
After six or seven generations, the selective breeding seems to have
reached a plateau. This suggests that the tomato plants have become monomorphic for the
alleles that affect tomato weight.
B.
There does seem to be heterosis, because the first generation has a weight
of 1.7 lb, which is heavier than either Mary’s or Hector’s tomatoes. This partially
explains why Martin has obtained tomatoes that are heavier than 1.5 lb. However,
heterosis is not the whole story; it does not explain why Martin obtained tomatoes that
weigh 2 lb. Even though Mary’s and Hector’s tomatoes were selected for heavier weight,
they may not have all of the “heavy alleles” for each gene that controls weight. For
example, let’s suppose there are 20 genes that affect weight, with each gene existing in a
light and heavy allele. During the early stages of selective breeding, when Mary and
Hector picked their 10 plants as seed producers for the next generation, as a matter of
random chance, some of these plants may have been homozygous for the light alleles at a
few of the 20 genes that control weight. Therefore, just as a matter of chance, they
probably “lost” a few of the heavy alleles that affect weight. So, after 12 generations of
breeding, they have predominantly heavy alleles but also have light alleles for some of
the genes. If we represent heavy alleles with a capital letter, and light alleles with a
lowercase letter, Mary’s and Hector’s strains could be the following:
Mary’s strain: AA BB cc DD EE FF gg hh II JJ KK LL mm NN OO PP QQ RR ss
TT
Hector’s strain:AA bb CC DD EE ff GG HH II jj kk LL MM NN oo PP QQ RR SS
TT
As we see here, Mary’s strain is homozygous for the heavy allele at 15 of the genes but
carries the light allele at the other 5. Similarly, Hector’s strain is homozygous for the
heavy allele at 15 genes and carries the light allele at the other 5. It is important to note,
however, that the light alleles in Mary’s and Hector’s strains are not in the same genes.
Therefore, when Martin crosses them together, he will initially get
Martin’s F1 offspring: AA Bb Cc DD EE Ff Gg Hh II Jj Kk LL Mm NN Oo PP
QQ RR Ss TT
If the alleles are additive and contribute equally to the trait, we would expect about the
same weight (1.5 lb), because this hybrid has a total of 10 light alleles. However, if
heterosis is occurring, genes (which were homozygous recessive in Mary’s and Hector’s
strains) will become heterozygous in the F1 offspring, and this may make the plants
healthier and contribute to a higher weight. If Martin’s F1 strain is subjected to selective
breeding, the 10 genes that are heterozygous in the F1 offspring may eventually become
homozygous for the heavy allele. This would explain why Martin’s tomatoes achieved a
weight of 2.0 pounds after five generations of selective breeding.
E13. The correlations for height were determined for 15 pairs of individuals with the
following genetic relationships:
Mother/daughter: 0.36
Mother/granddaughter: 0.17
Sister/sister: 0.39
Sister/sister (fraternal twins): 0.40
Sister/sister (identical twins): 0.77
What is the average heritability for height in this group of females?
Answer: hN2 = robs/rexp
The value for rexp comes from the known genetic relationships:
Mother/daughter
Mother/granddaughter
Sister/sister
Twin sisters (fraternal)
Twin sisters (identical)
robs = 0.36
robs = 0.17
robs = 0.39
robs = 0.40
robs = 0.77
rexp = 0.5
rexp = 0.25
rexp = 0.5
rexp = 0.5
rexp = 1.0
hN2 = 0.72
hN2 = 0.68
hN2 = 0.78
hN2 = 0.80
hN2 = 0.77
0.75
The average heritability is 0.75.
E14. An animal breeder had a herd of sheep with a mean weight of 254 lb at 3 years of
age. He chose animals with mean weights of 281 lb as parents for the next generation.
When these offspring reached 3 years of age, their mean weights were 269 lb.
A.
Calculate the narrow-sense heritability for weight in this herd.
B.
Using the heritability value that you calculated in part A, what weight of
animals would you have to choose to get offspring that weigh 275 lb (at 3 years of age)?
Answer:
A.
hN 2 
XO  X
XP  X
hN 2 
269  254
 0.56
281  254
B.
0.56 
275  254
X P  254
X P  291.5 lb
E15. The trait of blood pressure in humans has a frequency distribution that is similar
to a normal distribution. The following graph shows the ranges of blood pressures for a
selected population of people. The red line depicts the frequency distribution of the
systolic pressures for the entire population. Several individuals with high blood pressure
were identified, and the blood pressures of their relatives were determined. This
frequency distribution is depicted with a blue line. (Note: The blue line does not include
the people who were identified with high blood pressure; it includes only their relatives.)
[Insert Text Art 26.3]
What do these data suggest with regard to a genetic basis for high blood pressure? What
statistical approach could you use to determine the heritability for this trait?
Answer: These data suggest there might be a genetic component to blood pressure,
because the relatives of people with high blood pressure also seem to have high blood
pressure themselves. Of course, more extensive studies would need to be conducted to
determine the role of environment. To calculate heritability, the first thing to do is to
calculate the correlation coefficient between relatives to see if it is statistically
significant. If it is, then you could follow the approach described in the experiment of
Figure 26.8. You would determine the correlation coefficients between genetically related
individuals as a way to determine the heritability for the trait. In this approach,
heritability equals robs/rexp. It would be important to include genetically related pairs
that were raised apart (e.g., uncles and nieces) to see if they had a similar heritability
value compared to genetically related pairs raised in the same environment (e.g., brothers
and sisters). If their values were similar, this would give you some confidence that the
heritability value is due to genetics and not due to fact that relatives often share similar
environments.