1
The Genetics Revolution
PROBLEMS
In each chapter, a set of problems tests the reader’s comprehension of the concepts in the chapter and
their relation to concepts in previous chapters. Each problem set begins with some problems based on
the figures in the chapter, which embody important concepts. These are followed by problems of a more
general nature.
WORKING WITH THE FIGURES
1. If the white-flowered parental variety in Figure 1-3 were crossed to the first-generation hybrid plant
in that figure, what types of progeny would you expect to see and in what proportions?
Answer: You would get a 1:1 ratio of purple to white. This is because the first-generation hybrid
plant has one copy of the purple allele and one copy of the white allele, and as a result, 50 percent of
the gametes would carry the purple allele and 50 percent of the gametes would carry the white allele.
The white-flowered parental variety has two copies of the white allele, and all the gametes produced
from the white plant will carry the white allele. Hence, a cross between the two would produce a 1:1
ratio of purple to white.
Hybrid plant P/p ¥ white plant p/p
Gametes 50% P 50% p ¥ 100% p
50% P/p : 50% p/p
Purple : white
2. In Mendel’s 1866 publication as shown in Figure 1-4, he reports 705 purple (violet) flowered
offspring and 224 white-flowered offspring. The ratio he obtained is 3.15:1 for purple:white. How
do you think he explained the fact that the ratio is not exactly 3:1?
Answer: This depends on the sample size. When the sample size was large, the proportions were
close to 3:1 (e.g., for round and wrinkled seeds the ratio was 2.95:1 and the total population size
1
IGA 11e SM Ch 01.indd 1 11/12/14 2:05 PM
, 2 CHAPTER 1 The Genetics Revolution
was 7324), whereas for a small sample size such as the purple and white petal flowered plants (929
plants), the ratio was not as close to 3:1.
3. In Figure 1-6, the students have 1 of 15 different heights plus there are two height classes (4 ft 11 in
and 5 ft 0 in) for which there are no observed students. That is a total of 17 height classes. If a single
Mendelian gene can only account for two classes of a trait (such as purple or white flowers), how
many Mendelian genes would be minimally required to explain the observation of 17 height classes?
Answer: If a single gene can only account for two classes of a trait, minimum of 9 genes are required
to explain the 17 height classes.
4. Figure 1-7 shows a simplified pathway for arginine synthesis in Neurospora. Suppose you have a
special strain of Neurospora that makes citrulline but not arginine. Which gene(s) are likely mutant or
missing in your special strain? You have a second strain of Neurospora that makes neither citrulline
nor arginine but does make ornithine. Which gene(s) are mutant or missing in this strain?
Answer: If the mutant strain makes citrulline, that means genes A and B must be functional. Therefore,
the only gene that is missing or mutant in the first Neurospora strain must be gene C.
In the second strain, gene A must be functional since it is able to make ornithine. Gene B must be
missing or mutant since it is unable to make citrulline. However, gene C may or may not be missing/
mutant. Enzyme C converts citrulline into arginine (they are in the same sequential pathway), and
enzyme C is dependent on the availability of citrulline for its function.
5. Consider Figure 1-8a.
a. What do the small blue spheres represent?
b. What do the brown slabs represent?
c. Do you agree with the analogy that DNA is structured like a ladder?
Answer:
a. The blue ribbon represents sugar phosphate backbone (deoxyribose and a phosphate group), while
the blue spheres signify atoms.
b. Brown slabs show complementary bases (A, T, G, and C).
c. Yes, it is a helical structure.
6. In Figure 1-8b, can you tell if the number of hydrogen bonds between adenine and thymine is the
same as that between cytosine and guanine? Do you think that a DNA molecule with a high content
of A + T would be more stable than one with a high content of G + C?
Answer: There are two hydrogen bonds between adenine and thymine; three between guanine and
cytosine. No, the molecule with a high content of G-C would be more stable.
IGA 11e SM Ch 01.indd 2 11/12/14 2:05 PM
, CHAPTER 1 The Genetics Revolution 3
7. Which of three major groups (domains) of life in Figure 1-11 is not represented by a model organism?
Answer: Archaea
8. Figure 1-13b shows the human chromosomes in a single cell. The green dots show the location of a
gene called BAPX1. Is the cell in this figure a sex cell (gamete)? Explain your answer.
Answer: It is not a sex cell. Cloned BAPXI gene has hybridized to two chromosomes in the cell,
indicating there are two copies of the BAPXI gene. If it were a gamete, it would have only one copy
of each chromosome and of the BAPXI gene.
9. Figure 1-15 shows the family tree or pedigree for Louise Benge (Individual VI-1) who suffers from
the disease ACDC because she has two mutant copies of the CD73 gene. She has four siblings (VI-
2, VI-3, VI-4, and VI-5) who have this disease for the same reason. Do all the 10 children of Louise
and her siblings have the same number of mutant copies of the CD73 gene, or might this number be
different for some of the 10 children?
Answer: All 10 children have one mutant copy of the CD73 gene. Children get one CD73 copy from
their mom and one from their dad. Since Louise and her four siblings each carry two defective genes,
all their children will get one mutant CD73 copy.
BASIC QUESTIONS
10. Below is the sequence of a single strand of a short DNA molecule. On a piece of paper, rewrite this
sequence and then write the sequence of the complementary strand below it.
GTTCGCGGCCGCGAAC
Comparing the top and bottom strands, what do you notice about the relationship between them?
Answer:
GTTCGCGGCCGCGAAC
CAAGCGCCGGCGCTTG
They are complementary to each other and run in the opposite direction (antiparallel). The sequences
are also palindromic; they read the same in either direction.
11. Mendel studied a tall variety of pea plants with stems that are 20 cm long and a dwarf variety with
stems that are only 12 cm long.
a. Under blending theory, how long would you expect the stems of first and second hybrids to be?
b. Under Mendelian rules, what would you expect to observe in the second-generation hybrids if all
the first-generation hybrids were tall?
IGA 11e SM Ch 01.indd 3 11/12/14 2:05 PM
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Answer:
a. First-generation hybrids would have stems 16 cm long because that is the average between 20 cm
and 12 cm. The second generation would be a product of two 16-cm stemmed plants mating, so
they would also have stems 16 cm long.
b. If the first-generation hybrids are all tall, then tall must be dominant, and you would expect a 3:1
ratio of tall:dwarf in the second generation.
12. If a DNA double helix that is 100 base pairs in length has 32 adenines, how many cytosines, guanines,
and thymines must it have?
Answer: Thymines = 32, Cytosines = 68, Guanines = 68
Number of thymines is 32 because thymine and adenine are paired, so the number of adenines equals
the number of thymines. The remaining 136 base pairs must be cytosines and guanines. The number
of cytosines equals the number of guanines because those bases are paired. Therefore, there are 136
÷ 2 = 68 base pairs of each.
13. The complementary strands of DNA in the double helix are held together by hydrogen bonds: G≡C
or A=T. These bonds can be broken (denatured) in aqueous solutions by heating to yield two single
strands of DNA (see Figure 1-13a). How would you expect the relative amounts of GC versus AT
base pairs in a DNA double helix to affect the amount of heat required to denature it? How would
you expect the length of a DNA double helix in base pairs to affect the amount of heat required to
denature it?
Answer: Double-stranded DNA with a high GC content would be more stable because GC pairs
have three hydrogen bonds and hence would require more heat to denature compared with a double-
stranded DNA with high AT content.
As the length of the double helix increases, the heat required to denature would also increase. This is
because there would be more hydrogen bonds to break.
14. The figure that follows shows the DNA sequence of a portion of one of the chromosomes from a trio
(mother, father, and child). Can you spot any new point mutations in the child that are not in either
parent? In which parent did the mutation arise?
IGA 11e SM Ch 01.indd 4 11/12/14 2:05 PM
The Genetics Revolution
PROBLEMS
In each chapter, a set of problems tests the reader’s comprehension of the concepts in the chapter and
their relation to concepts in previous chapters. Each problem set begins with some problems based on
the figures in the chapter, which embody important concepts. These are followed by problems of a more
general nature.
WORKING WITH THE FIGURES
1. If the white-flowered parental variety in Figure 1-3 were crossed to the first-generation hybrid plant
in that figure, what types of progeny would you expect to see and in what proportions?
Answer: You would get a 1:1 ratio of purple to white. This is because the first-generation hybrid
plant has one copy of the purple allele and one copy of the white allele, and as a result, 50 percent of
the gametes would carry the purple allele and 50 percent of the gametes would carry the white allele.
The white-flowered parental variety has two copies of the white allele, and all the gametes produced
from the white plant will carry the white allele. Hence, a cross between the two would produce a 1:1
ratio of purple to white.
Hybrid plant P/p ¥ white plant p/p
Gametes 50% P 50% p ¥ 100% p
50% P/p : 50% p/p
Purple : white
2. In Mendel’s 1866 publication as shown in Figure 1-4, he reports 705 purple (violet) flowered
offspring and 224 white-flowered offspring. The ratio he obtained is 3.15:1 for purple:white. How
do you think he explained the fact that the ratio is not exactly 3:1?
Answer: This depends on the sample size. When the sample size was large, the proportions were
close to 3:1 (e.g., for round and wrinkled seeds the ratio was 2.95:1 and the total population size
1
IGA 11e SM Ch 01.indd 1 11/12/14 2:05 PM
, 2 CHAPTER 1 The Genetics Revolution
was 7324), whereas for a small sample size such as the purple and white petal flowered plants (929
plants), the ratio was not as close to 3:1.
3. In Figure 1-6, the students have 1 of 15 different heights plus there are two height classes (4 ft 11 in
and 5 ft 0 in) for which there are no observed students. That is a total of 17 height classes. If a single
Mendelian gene can only account for two classes of a trait (such as purple or white flowers), how
many Mendelian genes would be minimally required to explain the observation of 17 height classes?
Answer: If a single gene can only account for two classes of a trait, minimum of 9 genes are required
to explain the 17 height classes.
4. Figure 1-7 shows a simplified pathway for arginine synthesis in Neurospora. Suppose you have a
special strain of Neurospora that makes citrulline but not arginine. Which gene(s) are likely mutant or
missing in your special strain? You have a second strain of Neurospora that makes neither citrulline
nor arginine but does make ornithine. Which gene(s) are mutant or missing in this strain?
Answer: If the mutant strain makes citrulline, that means genes A and B must be functional. Therefore,
the only gene that is missing or mutant in the first Neurospora strain must be gene C.
In the second strain, gene A must be functional since it is able to make ornithine. Gene B must be
missing or mutant since it is unable to make citrulline. However, gene C may or may not be missing/
mutant. Enzyme C converts citrulline into arginine (they are in the same sequential pathway), and
enzyme C is dependent on the availability of citrulline for its function.
5. Consider Figure 1-8a.
a. What do the small blue spheres represent?
b. What do the brown slabs represent?
c. Do you agree with the analogy that DNA is structured like a ladder?
Answer:
a. The blue ribbon represents sugar phosphate backbone (deoxyribose and a phosphate group), while
the blue spheres signify atoms.
b. Brown slabs show complementary bases (A, T, G, and C).
c. Yes, it is a helical structure.
6. In Figure 1-8b, can you tell if the number of hydrogen bonds between adenine and thymine is the
same as that between cytosine and guanine? Do you think that a DNA molecule with a high content
of A + T would be more stable than one with a high content of G + C?
Answer: There are two hydrogen bonds between adenine and thymine; three between guanine and
cytosine. No, the molecule with a high content of G-C would be more stable.
IGA 11e SM Ch 01.indd 2 11/12/14 2:05 PM
, CHAPTER 1 The Genetics Revolution 3
7. Which of three major groups (domains) of life in Figure 1-11 is not represented by a model organism?
Answer: Archaea
8. Figure 1-13b shows the human chromosomes in a single cell. The green dots show the location of a
gene called BAPX1. Is the cell in this figure a sex cell (gamete)? Explain your answer.
Answer: It is not a sex cell. Cloned BAPXI gene has hybridized to two chromosomes in the cell,
indicating there are two copies of the BAPXI gene. If it were a gamete, it would have only one copy
of each chromosome and of the BAPXI gene.
9. Figure 1-15 shows the family tree or pedigree for Louise Benge (Individual VI-1) who suffers from
the disease ACDC because she has two mutant copies of the CD73 gene. She has four siblings (VI-
2, VI-3, VI-4, and VI-5) who have this disease for the same reason. Do all the 10 children of Louise
and her siblings have the same number of mutant copies of the CD73 gene, or might this number be
different for some of the 10 children?
Answer: All 10 children have one mutant copy of the CD73 gene. Children get one CD73 copy from
their mom and one from their dad. Since Louise and her four siblings each carry two defective genes,
all their children will get one mutant CD73 copy.
BASIC QUESTIONS
10. Below is the sequence of a single strand of a short DNA molecule. On a piece of paper, rewrite this
sequence and then write the sequence of the complementary strand below it.
GTTCGCGGCCGCGAAC
Comparing the top and bottom strands, what do you notice about the relationship between them?
Answer:
GTTCGCGGCCGCGAAC
CAAGCGCCGGCGCTTG
They are complementary to each other and run in the opposite direction (antiparallel). The sequences
are also palindromic; they read the same in either direction.
11. Mendel studied a tall variety of pea plants with stems that are 20 cm long and a dwarf variety with
stems that are only 12 cm long.
a. Under blending theory, how long would you expect the stems of first and second hybrids to be?
b. Under Mendelian rules, what would you expect to observe in the second-generation hybrids if all
the first-generation hybrids were tall?
IGA 11e SM Ch 01.indd 3 11/12/14 2:05 PM
, 4 CHAPTER 1 The Genetics Revolution
Answer:
a. First-generation hybrids would have stems 16 cm long because that is the average between 20 cm
and 12 cm. The second generation would be a product of two 16-cm stemmed plants mating, so
they would also have stems 16 cm long.
b. If the first-generation hybrids are all tall, then tall must be dominant, and you would expect a 3:1
ratio of tall:dwarf in the second generation.
12. If a DNA double helix that is 100 base pairs in length has 32 adenines, how many cytosines, guanines,
and thymines must it have?
Answer: Thymines = 32, Cytosines = 68, Guanines = 68
Number of thymines is 32 because thymine and adenine are paired, so the number of adenines equals
the number of thymines. The remaining 136 base pairs must be cytosines and guanines. The number
of cytosines equals the number of guanines because those bases are paired. Therefore, there are 136
÷ 2 = 68 base pairs of each.
13. The complementary strands of DNA in the double helix are held together by hydrogen bonds: G≡C
or A=T. These bonds can be broken (denatured) in aqueous solutions by heating to yield two single
strands of DNA (see Figure 1-13a). How would you expect the relative amounts of GC versus AT
base pairs in a DNA double helix to affect the amount of heat required to denature it? How would
you expect the length of a DNA double helix in base pairs to affect the amount of heat required to
denature it?
Answer: Double-stranded DNA with a high GC content would be more stable because GC pairs
have three hydrogen bonds and hence would require more heat to denature compared with a double-
stranded DNA with high AT content.
As the length of the double helix increases, the heat required to denature would also increase. This is
because there would be more hydrogen bonds to break.
14. The figure that follows shows the DNA sequence of a portion of one of the chromosomes from a trio
(mother, father, and child). Can you spot any new point mutations in the child that are not in either
parent? In which parent did the mutation arise?
IGA 11e SM Ch 01.indd 4 11/12/14 2:05 PM
