Introduction to Genetic Analysis Eleventh Edition Solution Manual

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1
1
The GeneTics R evoluTion
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
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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.
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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?
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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?
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Answer: There is a new point mutation from an A to a C. The child inherited copy F1 from the father
without any de novo mutations, as shown on the bottom. The other copy (as shown on the top) must
therefore be from the mother. The mother has an SNP (A/T) at the fifth nucleotide shown, so we know
the child inherited copy M1 (A). The A-to-C mutation in copy M1 is not observed in either of the
parents, so it must be a new mutation.
CHALLENGING PROBLEMS
15. a. There are three nucleotides in each codon, and each of these nucleotides can have one of four
different bases. How many possible unique codons are there?
b. If DNA had only two types of bases instead of four, how long would codons need to be to specify
all 20 amino acids?
Answer:
a. There are 64 unique codons (4 ¥ 4 ¥ 4 = 64).
b. In order to specify 20 amino acids using only two bases, a codon must be five bases long (2 ¥ 2 ¥
2 ¥ 2 ¥ 2 = 32). This would give 32 unique codons that could specify 20 amino acids.
16. Fathers contribute more new point mutations to their children than mothers. You may know from
general biology that people have sex chromosomes—two X chromosomes in females and an X plus
a Y chromosome in males. Both sexes have the autosomes (As).
a. On which type of chromosome (A, X, or Y) would you expect the genes to have the greatest
number of new mutations per base pair over many generations in a population? Why?
b. On which type of chromosome would you expect the least number of new mutations per base
pair? Why?
c. Can you calculate the expected number of new mutations per base pair for a gene on the X and Y
chromosome for every 1 new mutation in a gene on an autosome if the mutation rate in males is
twice that in females?
Answer:
a. Y chromosome. All, or 100 percent of, Y chromosomes are passed through the male lineage via
sperm. The production of sperm (spermatogenesis) continues throughout a man’s life, undergoing
many rounds of cell division. Each round of cell division include DNA replication, which provides
opportunity for new mutations, so the male lineage has a higher mutation rate than the female
lineage, as explained in the text.
b. X chromosome. Two out of every three X chromosomes (66.7 percent) are transmitted through the
female lineage, which has a lower mutation rate than the male lineage. By comparison, 50 percent
of autosomes are inherited through females with the lower mutation rate and 50 percent through
males with the higher mutation rate.
c. Set the male mutation rate (mM) as 2.0 per unit time and the female mutation rate (mF ) as 1.0 per
unit time. Since autosomes are transmitted equally through males and females, then the autosomal
rate is the average of the male and female rates:
(mA) = (2 + 1)/2 = 1.5
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Since genes on the Y are transmitted only through males, the Y rate equals the male rate:
mY = mM = 2
Since 1/3 of X chromosomes are inherited through males and 2/3 through females:
mX = (1/3 mM) + (2/3 mF ) = (1/3 ¥ 2) + (2/3 ¥ 1) = 1.33
The expected number of new mutation on the Y for every 1 on an autosome is 2/1.5 = 1.33. The
expected number of new mutation on the X for every 1 on an autosome is 1.33/1.5 = 0.889.
17. For young men of age 20, there have been 150 rounds of DNA replication during sperm production
as compared with only 23 rounds for a woman of age 20. That is a 6.5-fold greater number of cell
divisions and proportionately greater opportunity for new point mutations. Yet, on average, 20-year-
old men contribute only about twice as many new point mutations to their offspring as do women.
How can you explain this discrepancy?
Answer: While experimental evidence to explain this observation are not available, one hypothesis is
that sperm cells are physiologically weak and their normal function is easily disrupted by mutations.
Thus, sperm that carry deleterious new mutations are less likely to survive and form a zygote with
an egg cell. Therefore, many sperm with new mutations are eliminated prezygotically.
18. In computer science, a bit stores one of two states, 0 or 1. A byte is a group of 8 bits, which has 2 8
= 256 possible states. Modern computer files are often mega-bytes (106 bytes), or even giga-bytes
(109 bytes), in size. The human genome is approximately 3 billion base pairs in size. How many
nucleotides are needed to encode a single byte? How large of a computer file would it take to store
the same amount of information as a single human genome?
Answer: Four nucleotides, with four possible states (A, T, C, G) can encode 4 ¥ 4 ¥ 4 ¥ 4 = 44 = 256
= 1 byte. If one byte can encode four nucleotides, then 3 ¥ 109 nucleotides can be encoded in (3 ¥
10 9 ) / 4 = 7.5 ¥ 108 bytes = 750 ¥ 106
, or 750 megabytes.
19. The human genome is approximately 3 billion base pairs in size.
a. Using standard 8.5-in ¥ 11-in paper with one-inch margins, a 12-point font size and single-spaced
lines, how many sheets of paper printed on one side would be required to print out the human
genome?
b. A ream of 500 sheets of paper is about 5 cm thick. How tall would the stack of paper with the
entire human genome be?
c. Would you want a backpack, shopping cart, or a semi trailer truck to haul around this stack?
Answer:
a. Assuming a single one-sided page can fit 23 lines, with 56 letters in each line, 23 ¥ 56 = 1288
letters per sheet. The human genome is 3 ¥ 109 base pairs, so the number of sheets required to print
out the entire human genome = 3 ¥ 10 9 /1288 = 2,329,192 sheets of paper.
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8
2
Single-gene inheritance
WORKING WITH THE FIGURES
1. In the left-hand part of Figure 2-4, the red arrows show selfing as pollination within single flowers
of one F 1 plant. Would the same F2 results be produced by cross-pollinating two different F 1 plants?
Answer: No, the results would be different. While self-pollination produces 3:1 ratio of yellow ver-
sus gene phenotype, cross-pollination would result in 1:1 ratio in the F 2 . This is because F 1 yellow
are heterozygous, while green are homozygous genotypes.
2. In the right-hand part of Figure 2-4, in the plant showing an 11:11 ratio, do you think it would be
possible to find a pod with all yellow peas? All green? Explain.
Answer: Yes, it is possible to find complete pods with only yellow peas or only green peas from the
cross shown, though it would be highly unlikely. It would have the same likelihood of occurring as
flipping a coin and getting heads six times in a row.
3. In Table 2-1, state the recessive phenotype in each of the seven cases.
Answer: wrinkled seeds; green seeds; white petals; pinched pods; yellow pods; terminal flowers;
short stems
4. Considering Figure 2-8, is the sequence “pairing → replication → segregation → segregation” a
good shorthand description of meiosis?
Answer: No, it should say either “pairing, recombination, segregation, segregation” or “replication,
pairing, segregation, segregation.”
5. Point to all cases of bivalents, dyads, and tetrads in Figure 2-11.
Answer: Replicate sister chromosomes or dyads are at any chromatid after the replication (S phase).
A pair of synapsed dyads is called a bivalent, and it would represent two dyads together (sister
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chromatids on the right), while the four chromatids that make up a bivalent are called a tetrad, and
they would be the entire square (with same or different alleles on the bivalents).
6. In Figure 2-11, assume (as in corn plants) that allele A encodes an allele that produces starch in pol-
len and allele a does not. Iodine solution stains starch black. How would you demonstrate Mendel’s
first law directly with such a system?
Answer: One would use this iodine dye to color the starch-producing corn pollen. Since pollen is a
plant gametophyte generation (haploid), it will be produced by meiosis. Mendel’s first law predicts
segregation of alleles into gametes; therefore, we would expect 1:1 ratio of starch-producing (A) ver-
sus non-starch-producing (a) pollen grains, from a heterozygous (A/a) parent/male flower. It would
be easy to color the pollen and count the observed ratio.
7. Considering Figure 2-13, if you had a homozygous double mutant m3/m3 m5/m5, would you ex-
pect it to be mutant in phenotype? (Note: This line would have two mutant sites in the same coding
sequence.)
Answer: Yes, this double mutant m3/m3 m5/m5 would be a null mutation because m3 mutation
changes the exon sequence. Because the m5 mutation is silent, the homozygous double mutant
m3/m3 m5/m5 would have the same mutant phenotype as an m3/m3 double mutant.
8. In which of the stages of the Drosophila life cycle (represented in the box on page 56) would you
find the products of meiosis?
Answer: Meiosis happens in adult ovaries and testes to produce gametes (sperm and unfertilized
egg). Thus, the adult fly in the diagram would generate gametes and participate in mating, and the
female would then lay the fertilized diploid embryos (eggs).
9. If you assume Figure 2-15 also applies to mice and you irradiate male sperm with X rays (known to
inactivate genes via mutation), what phenotype would you look for in progeny in order to find cases
of individuals with an inactivated SRY gene?
Answer: Individuals with an inactivated SRY gene would be phenotypically female with an XY sex
chromosomal makeup. These individuals are often called “sex reversed” and are always sterile.
Hence, we would look for flies that are phenotypically female, but sterile.
10. In Figure 2-17, how does the 3:1 ratio in the bottom-left-hand grid differ from the 3:1 ratios obtained
by Mendel?
Answer: It differs because, in Mendel’s experiments, we learned about autosomal genes, while in
this case, we have a sex-linked gene for eye color.
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3:1 ratio means that all females have red eyes (X +/–
), while half the males have red (X +
/Y) and half
have white (XW/Y). Careful sex determination when counting F2 offspring would point out to a sex-
linked trait.
11. In Figure 2-19, assume that the pedigree is for mice, in which any chosen cross can be made. If you
bred IV-1 with IV-3, what is the probability that the first baby will show the recessive phenotype?
Answer: 2/3 ¥ 2/3 ¥ 1/4 = 1/9, or 0.11
The probability that IV-1 and IV-3 mice are heterozygous is 2/3. This is because both of their parents
are known heterozygotes (A/a), and since they are the dominant phenotype, they could only be A/A
or A/a. Now the probability that two heterozygotes have a recessive homozygote offspring is 1/4.
12. Which part of the pedigree in Figure 2-23 in your opinion best demonstrates Mendel’s first law?
Answer: Any part of this pedigree demonstrates the law, showing segregation of alleles into gametes.
Notice how 50 percent of the children of I-1 and I-2 display the dominant trait, consistent with I-1
contributing either the dominant or the recessive allele to each gamete in equal frequencies (1:1).
The middle part of generation II marriage shows a typical testcross (expected 1:1). Neither ratio in
the pedigree could be confirmed because of a small sample size in any given family, but allele seg-
regation is obvious.
13. Could the pedigree in Figure 2-31 be explained as an autosomal dominant disorder? Explain.
Answer: Yes, it could in some cases, but in this case we have clues that the pedigree is for a sex-
linked dominant trait. First, if fathers have a gene, only daughters would receive it; and second, if
mothers have a gene, both sons and daughters would receive it.
BASIC PROBLEMS
14. Make up a sentence including the words chromosome, genes, and genome.
Answer: The human genome contains an estimated 20,000–25,000 genes located on 23 different
chromosomes.
15. Peas (Pisum sativum) are diploid and 2n = 14. In Neurospora, the haploid fungus, n = 7. If it were
possible to fractionate genomic DNA from both species by using pulsed field electrophoresis, how
many distinct DNA bands would be visible in each species?
Answer: PFGE separates DNA molecules by size. When DNA is carefully isolated from Neurospora
(which has seven different chromosomes), seven bands should be produced using this technique.
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Similarly, the pea has seven different chromosomes and will produce seven bands (homologous
chromosomes will co-migrate as a single band).
16. The broad bean (Vicia faba) is diploid and 2n = 18. Each haploid chromosome set contains approxi-
mately 4 m of DNA. The average size of each chromosome during metaphase of mitosis is 13 mm.
What is the average packing ratio of DNA at metaphase? (Packing ratio = length of chromosome/
length of DNA molecule therein.) How is this packing achieved?
Answer: There is a total of 4 m of DNA and nine chromosomes per haploid set. On average, each is
4/9 m long. At metaphase, their average length is 13 mm, so the average packing ratio is 13 ¥ 10–6
m:4.4 ¥ 10–1 m, or roughly 1:34,000! This remarkable achievement is accomplished through the
interaction of the DNA with proteins. At its most basic, eukaryotic DNA is associated with histones
in units called nucleosomes, and during mitosis, coils into a solenoid. As loops, it associates with and
winds into a central core of nonhistone protein called the scaffold.
17. If we call the amount of DNA per genome “x,” name a situation or situations in diploid organisms in
which the amount of DNA per cell is:
a. x b. 2x c. 4x
Answer: Because the DNA levels vary four-fold, the range covers cells that are haploid (gametes) to
cells that are dividing (after DNA has replicated but prior to cell division). The following cells would
fit the DNA measurements:
a. x haploid cells
b. 2x diploid cells in G1 or cells after meiosis I but prior to meiosis II
c. 4x diploid cells after S but prior to cell division
18. Name the key function of mitosis.
Answer: The key function of mitosis is to generate two daughter cells that are genetically identical
to the original parent cell.
19. Name two key functions of meiosis.
Answer: Two key functions of meiosis are to halve the DNA content and to reshuffle the genetic
content of the organism to generate genetic diversity among the progeny.
20. Design a different nuclear-division system that would achieve the same outcome as that of meiosis.
Answer: It’s pretty hard to beat several billion years of evolution, but it might be simpler if DNA
did not replicate prior to meiosis. The same events responsible for halving the DNA and producing
genetic diversity could be achieved in a single cell division if homologous chromosomes paired,
recombined, randomly aligned during metaphase, and separated during anaphase, etc. However, you
would lose the chance to check and repair DNA that replication allows.
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21. In a possible future scenario, male fertility drops to zero, but, luckily, scientists develop a way for
women to produce babies by virgin birth. Meiocytes are converted directly (without undergoing
meiosis) into zygotes, which implant in the usual way. What would be the short-term and long-term
effects in such a society?
Answer: In large part, this question is asking: Why sex? Parthenogenesis (the ability to reproduce
without fertilization—in essence, cloning) is not common among multicellular organisms. Parthe-
nogenesis occurs in some species of lizards and fishes and several kinds of insects, but it is the only
means of reproduction in only a few of these species. In plants, about 400 species can reproduce
asexually by a process called apomixis. These plants produce seeds without fertilization. However,
the majority of plants and animals reproduce sexually. Sexual reproduction produces a wide variety
of different offspring by forming new combinations of traits inherited from both the father and the
mother. Despite the numerical advantages of asexual reproduction, most multicellular species that
have adopted it as their only method of reproducing have become extinct. However, there is no
agreed-upon explanation of why the loss of sexual reproduction usually leads to early extinction, or
conversely, why sexual reproduction is associated with evolutionary success. On the other hand, the
immediate effects of such a scenario are obvious. All offspring would be genetically identical to their
mothers, and males would be extinct within one generation.
22. In what ways does the second division of meiosis differ from mitosis?
Answer: As cells divide mitotically, each chromosome consists of identical sister chromatids that are
separated to form genetically identical daughter cells. Although the second division of meiosis ap-
pears to be a similar process, the “sister” chromatids are likely to be different. Recombination during
earlier meiotic stages has swapped regions of DNA between sister and nonsister chromosomes such
that the two daughter cells of this division typically are not genetically identical.
23. Make up mnemonics for remembering the five stages of prophase I of meiosis and the four stages of
mitosis.
Answer: The four stages of mitosis are prophase, metaphase, anaphase, and telophase. The first let-
ters, PMAT, can be remembered by a mnemonic such as Playful Mice Analyze Twice.
The five stages of prophase I are leptotene, zygotene, pachytene, diplotene, and diakinesis. The first
letters, LZPDD, can be remembered by a mnemonic such as Large Zoos Provide Dangerous Distrac-
tions.
24. In an attempt to simplify meiosis for the benefit of students, mad scientists develop a way of pre-
venting premeiotic S phase and making do with having just one division, including pairing, crossing
over, and segregation. Would this system work, and would the products of such a system differ from
those of the present system?
Answer: Yes, it could work, but certain DNA repair mechanisms (such as postreplication recombina-
tion repair) could not be invoked prior to cell division. There would be just two cells as products of
this meiosis, rather than four.
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25. Theodor Boveri said, “The nucleus doesn’t divide; it is divided.” What was he getting at?
Answer: The nucleus contains the genome and separates it from the cytoplasm. However, during
cell division, the nuclear envelope dissociates (breaks down). It is the job of the microtubule-based
spindle to actually separate the chromosomes (divide the genetic material) around which nuclei
reform during telophase. In this sense, it can be viewed as a passive structure that is divided by the
cell’s cytoskeleton.
26. Francis Galton, a geneticist of the pre-Mendelian era, devised the principle that half of our genetic
makeup is derived from each parent, one-quarter from each grandparent, one-eighth from each great-
grandparent, and so forth. Was he right? Explain.
Answer: Yes, half of our genetic makeup is derived from each parent, each parent’s genetic makeup
is derived half from each of their parents, etc. The process is a bit more complex when one considers
the recombination of homologous chromosomes in prophase I, as is discussed in later chapters.
27. If children obtain half their genes from one parent and half from the other parent, why aren’t siblings
identical?
Answer: Because the “half” inherited is very random, the chances of receiving exactly the same half
is vanishingly small. Ignoring recombination and focusing just on which chromosomes are inherited
from one parent, there are 223 = 8,388,608 possible combinations!
28. State where cells divide mitotically and where they divide meiotically in a fern, a moss, a flowering
plant, a pine tree, a mushroom, a frog, a butterfly, and a snail.
Answer:
U Mitosis U Meiosis
fern sporophyte gametophyte (sporangium)
moss sporophyte
gametophyte
sporophyte (antheridium and archegonium)
plant sporophyte gametophyte sporophyte (anther and ovule)
pine tree sporophyte gametophyte sporophyte (pine cone)
mushroom sporophyte gametophyte sporophyte (ascus or basidium)
frog somatic cells gonads
butterfly somatic cells gonads
snail somatic cells gonads
29. Human cells normally have 46 chromosomes. For each of the following stages, state the number of
nuclear DNA molecules present in a human cell:
a. Metaphase of mitosis
b. Metaphase I of meiosis
c. Telophase of mitosis
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d. Telophase I of meiosis
e. Telophase II of meiosis
Answer: This problem is tricky because the answers depend on how a cell is defined. In general,
geneticists consider the transition from one cell to two cells to occur with the onset of anaphase in
both mitosis and meiosis, even though cytoplasmic division occurs at a later stage.
a. 46 chromosomes, each with two chromatids = 92 chromatids
b. 46 chromosomes, each with two chromatids = 92 chromatids
c. 46 physically separate chromosomes in each of two about-to-be-formed cells
d. 23 chromosomes in each of two about-to-be-formed cells, each with two chromatids =
46 chromatids
e. 23 chromosomes in each of two about-to-be-formed cells
30. Four of the following events are part of both meiosis and mitosis, but only one is meiotic. Which
one? (1) chromatid formation, (2) spindle formation, (3) chromosome condensation, (4) chromo-
some movement to poles, (5) synapsis
Answer: (5) chromosome pairing (synapsis)
31. In corn, the allele ƒ´ causes floury endosperm and the allele f´´ causes flinty endosperm. In the cross
ƒ´/ƒ´ ♀ ¥ ƒ´´/ƒ´´ ♂, all the progeny endosperms are floury, but in the reciprocal cross, all the progeny
endosperms are flinty. What is a possible explanation? (Check the legend for Figure 2-7.)
Answer: First, examine the crosses and the resulting genotypes of the endosperm:
Female Male Polar nuclei Sperm Endosperm
ƒ´/ƒ´ ƒ´´/ƒ´´ ƒ´ and ƒ´ ƒ´´/ƒ´´ ƒ´/ƒ´/ƒ´´ (floury)
ƒ´´/ƒ´´ ƒ´/ƒ´ ƒ´´ and ƒ´´ ƒ´/ƒ´ ƒ´´/ƒ´´/ƒ´ (flinty)
As can be seen, the phenotype of the endosperm correlates to the predominant allele present.
32. What is Mendel’s first law?
Answer: Mendel’s first law states that alleles segregate into gametes during meiosis. This discovery
came from his monohybrid experimental crosses.
33. If you had a fruit fly (Drosophila melanogaster) that was of phenotype A, what test would you make
to determine if the fly’s genotype was A/A or A/a?
Answer: Do a testcross (cross to a/a). If the fly was A/A, all the progeny will be phenotypically A; if
the fly was A/a, half the progeny will be A and half will be a.
14CHAPTER 2 Single-Gene Inheritance
IGA 11e SM Ch 02.indd 14 11/12/14 2:30 PM

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34. In examining a large sample of yeast colonies on a petri dish, a geneticist finds an abnormal-looking
colony that is very small. This small colony was crossed with wild type, and products of meiosis
(ascospores) were spread on a plate to produce colonies. In total, there were 188 wild-type (normal-
size) colonies and 180 small ones.
a. What can be deduced from these results regarding the inheritance of the small-colony pheno-
type? (Invent genetic symbols.)
b. What would an ascus from this cross look like?
Answer:
a. A diploid meiocyte that is heterozygous for one gene (for example, s+
/s, where s is the allele that
confers the small colony phenotype) will, after replication and segregation, give two meiotic
products of genotype s+ and two of s. If the random spores of many meiocytes are analyzed, you
would expect to find about 50 percent normal-size colonies and 50 percent small colonies if the
abnormal phenotype is the result of a mutation in a single gene. Thus, the actual results of 188
normal-size and 180 small-size colonies support the hypothesis that the phenotype is the result
of a mutation in a single gene.
b. The following represents an ascus with four spores. The important detail is that two of the spores
are s and will generate small colonies, and two are s+ and will generate normal colonies.
35. Two black guinea pigs were mated and over several years produced 29 black and 9 white offspring.
Explain these results, giving the genotypes of parents and progeny.
Answer: The progeny ratio is approximately 3:1, indicating classic heterozygous-by-heterozygous
mating. Since black (B) is dominant to white (b):
Parents: B/b ¥ B/b
Progeny: 3 black:1 white (1 B/B:2 B/b:1 b/b)
This ratio indicates that black parents were probably heterozygous and that black is dominant over
white.
36. In a fungus with four ascospores, a mutant allele lys-5 causes the ascospores bearing that allele to be
white, whereas the wild-type allele lys-5+ results in black ascospores. (Ascospores are the spores that
constitute the four products of meiosis.) Draw an ascus from each of the following crosses:
a. lys-5 ¥ lys-5+
b. lys-5 ¥ lys-5
c. lys-5+ ¥ U lysU-5+
15CHAPTER 2 Single-Gene Inheritance
IGA 11e SM Ch 02.indd 15 11/12/14 2:30 PM

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Introduction to Genetic Analysis Eleventh Edition Solution Manual

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