TEST BANK FOR Introduction to Genetic Analysis, 10th Edition By Anthony J.F. Griffiths
- Rating : 0
- Grade : No Rating
- Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00
The Genetics Revolution in the Life
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. In considering Figure 1-2, if you were to extend the diagram, what would the next
two stages of “magnification” beyond DNA be?
Answer: The next stage of the diagram would be an in-depth look into the DNA
molecule, as two long molecular strands of nucleotides wound around each other in
a double helix and the basic structure of those monomers. Focus on specific
nucleotides and their organic molecule parts: a deoxyribose sugar, a phosphate
group, and a nitrogenous base (adenine, thymine, guanine, and cytosine). Second,
would be atomic composition of those organic molecules (C, H, O, P, and N).
2. In considering Figure 1-3,
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?
a. 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.
2 Chapter One
3. In Figure 1-4, 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
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.
4. From Figure 1-6, can you predict how many chromosomes there would be in a
muntjac sperm? How many purple chromosomes would there be in a sperm cell?
Answer: There would be only three chromosomes in a sperm cell of this species.
Since each homologous chromosome pair is stained with a different color, there
would be only one purple chromosome in a sperm cell.
5. In examining Figure 1-7, state one major difference between the chromosomal
“landscapes” of yeast and Drosophila.
Answer: Yeast chromosome landscape shows fewer introns and less space between
the coding genes.
6. In Figure 1-8, is it true that the direction of transcription is from right to left as
written for all the genes shown in these chromosomal segments?
Answer: No, there is one gene that would be read from left to right, since RNA
polymerase can assemble polynucleotides only in the 5 -3 direction.
7. In Figure 1-9, estimate what length of DNA is shown in the right-hand part of the
Answer: The right-hand part of this figure shows a section of a 30 nm fiber, which
is composed of nucleosomes (10 nm fibers) DNA alone is a 2 nm fiber. If
stretched out, a DNA molecule of each chromosome would be about 4 cm long,
thousands of times the diameter of a cell nucleus.
8. From Figure 1-12, what is the main difference in the locales of transcription and
Answer: In a eukaryotic cell the nucleus provides a separate location for
Chapter One 3
transcription, while translation continues in the cell cytoplasm.
9. In Figure 1-14, what do the colors blue and gold represent?
Answer: Blue represents original DNA (chromatid) in a cell before replication,
while gold represents new DNA (sister chromatids) after the semi-conservative
replication of the chromosome.
10. From Figure 1-17, locate the chromosomal positions of three genes involved in
tumor production in the human body.
Answer: There are many genes involved in tumor production in humans, such as a
gene for: neurofibromatosis on chromosome 2, familial colon cancer on
chromosome 2, for malignant melanoma on chromosome 9, and retinoblastoma on
11. In Figure 1-18, calculate the approximate number of nucleotide differences between
humans and dogs in the cytochrome c gene. Repeat for humans and moths.
Considering that the gene is several hundred nucleotides long, do these numbers
seem large or small to you? Explain.
Answer: Cytochrome c appears somewhat different when compared between
humans and dogs, since they diverged with approximately 14 nucleotide
substitutions since the common ancestor. Humans and moths differ even more, in
about 32 nucleotide substitutions, yet the difference is not as large as expected
based on the broad biological differences between insects and mammals. These
could tell us that the cytochrome c gene has been highly conserved due to its
significance in metabolism of aerobic organisms.
12. In Figure 1-21, why are colored ladders of bands shown in all three electrophoretic
gels? If the molecular labels used in all cases were radioactive, do you think the
black bands in the bottom part of the figure would all be radioactive?
Answer: In this figure we see three different types of electrophoresis in different
colors (Southern blot of DNA fragments, Northern blot of RNA and Western blot
of a protein product). Such a mixture of macromolecules could be hybridized with a
radioactive probe, and the bands in the lower part of the figure would indicate
13. In this chapter, the statement is made that most of the major questions of biology
4 Chapter One
have been answered through genetics. What are the main questions of biology, and
do you agree with the above statement? (State your reasons.)
Answer: Biological sciences inquire about life and its properties. Many themes
connect concepts and study life’s properties at the different levels of biological
hierarchy. One main theme is the continuity of life, which is based on heredity.
Genetics studies this theme in a great detail. Another theme is of course, evolution
of life, where again genetics plays a major role in understanding life history and
unity, as well as diversity of life.
14. It has been said that the DNA → RNA → protein discovery was the “Rosetta stone”
of biology. Do you agree?
Answer: Yes, this is the main aspect of the information processing in a cell: from
DNA to RNA and protein; from genotype to phenotype. Although this has been a
“central dogma of molecular biology” for decades, we know today that it has its
exceptions, such as the reverse transcription (RNA viruses) or the small RNA and
their role in gene regulation.
Understanding of this essential concept of life gives us an insight into another,
such as evolution, and the nature of mutations as a basic source of variability upon
which evolutionary processes might act.
15. Who do you think had the greatest impact on biology, Charles Darwin or the
research partners James Watson and Francis Crick?
Answer: Charles Darwin made an enormous impact on Biological sciences and
society, and his works are studied in many different areas, continuing to make an
impact today. A hundred years later, James Watson and Francis Crick discovered
the double helix, a molecule of life’s heritable information. This was perhaps the
most significant milestone in genetics and beyond. If Darwin had any information
about genes, even more about the properties of life’s blueprints, his theory would
have an important mechanism. It was the scientists of the first half of the twentieth
century who made a connection between the works of Darwin and Mendel in the
“great synthesis” and those in the second half of the twentieth century who made a
connection of all these milestones in the field of molecular evolution. Today, in the
twenty-first century both themes grew into studies of genomes and phylogenies and
at the even higher level into integrative and systems biology. It is hard to say whose
contribution is greater, but we must see their presence in the entire realm of the
16. How has genetics affected (a) agriculture, (b) medicine, (c) evolution, and (d)
modern biological research?
Chapter One 5
a. Genetics has affected agriculture for thousands of years, yet since the early
twentieth century this impact has been essential. Knowledge of the genetic basis of
traits and the experimental crossing allowed the growth in all of the fields of
agriculture. Besides artificial selection and breeding strategies, recombinant DNA
technology lead to genetic engineering and amazing results in this filed.
b. One of the fastest growing areas of genetics is the area involved with human
health and medicine. Genetics plays an essential role in studies of many diseases,
such as numerous hereditary diseases, cancer, diabetes, etc. Many genetic
disciplines are constantly involved in such studies and practices to understand and
diagnose human diseases. In addition, genetics plays an essential role in
reproductive biology. Finally, genetics might be used to cure diseases, whether
through gene therapy, stem cell treatments, or pharmacogenomics.
c. Evolution could be defined as a change in genetic makeup of a population over
time or, at a more broad level in Darwin’s words, as a descent with modification. In
the light of modern genetics, we could see how changes in genomes support the
concept that all of the living species descend from ancestral species. Evolution is
supported by an extensive amount of evidence, above all genetic evidence, which
continues to enrich our understanding of life’s unity and diversity. For
example, phylogenies constructed on genetic studies of species show evolutionary
relationships, enabling scientists to construct the tree of life. Besides such studies
based on genetics, studies in population, quantitative and developmental genetics,
molecular genetics, and bioinformatics bring new insights on evolution as a
unifying theory of all biology.
d. Genetics is the essential discipline in modern biological research and it is present
in almost every field of study. DNA cloning and polymerase chain reaction
techniques changed the way modern biology operates. At the same time, DNA
technology allows us to find genes of interest and study their function.
Reproductive cloning of mammals, genetic engineering, forensics, stem cell
research, diagnosis of human diseases, and gene therapy are new areas in
modern biology founded on genetics.
17. Assume for the sake of this question that the human body contains a trillion cells (a
low estimate). We know that a human genome contains about 1 meter of DNA. If
all the DNA in your body were laid end to end, do you think it could stretch to the
Moon and back? Justify your answer with a calculation. (Note: The average
distance to the Moon is 385,000 kilometers.)
Answer: Yes, if we could take DNA molecules from all of the nuclei in an
individual human and lay them straight, one after another, a total length of such
nucleic acid polymer would be equal: number of cells in a human body (trillion or
1,000,000,000,000) length of each cell’s DNA (1 m or 0.001 km) =
1,000,000,000 km, enough to reach the Moon and return. The key to such
enormous lengths is the chromosome packaging.
Single Gene Inheritance
WORKING WITH THE FIGURES
(The first 14 questions require inspection of text figures.)
1. In the left-hand part of Figure 2-4, the red arrows show selfing as pollination
within single flowers of one F1 plant. Would the same F2 results be produced by
cross-pollinating two different F1 plants?
Answer: No, the results would be different. While self pollination produces 3 : 1
ratio of yellow versus gene phenotype, cross pollination would result in 1 : 1
ratio, in the F2. This is because F1 yellow are heterozygous, while green are
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?
Answer: Yes, it is possible to find a pod with only yellow peas or heterozygous
for the seed color gene, if all the flowers had dominant allele in a given
fruit/pod. This could be also one example of rare changes at a physiological
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.”
Chapter Two 7
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 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 2-12, assume (as in corn plants) that A encodes an allele that produces
starch in pollen 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) versus
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. In the text figure on page 43, assume the left-hand individual is selfed. What
pattern of radioactive bands would you see in a Southern analysis of the
Answer: If an individual is selfed, the restriction fragments should be identical
to the parents fragments. In this case, a heterozygous parent to the left had three
bands (two from a mutant allele “a” and one from dominant allele “A”).
8. Considering Figure 2-15, if you had a homozygous double mutant m3/m3
m5/m5, would you expect 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 and m5/m5 would be a null mutation,
because m3 mutation changes the exon sequence.
9. In which of the stages of the Drosophila life cycle (represented in the box on
page 52) does meiosis take place?
Answer: Meiosis happens in adult ovaries and testes, therefore before
fertilization. After fertilization, fruit flies would lay their eggs (with now
diploid embryos). That would be Stage 1 on the figure.
8 Chapter Two
10. If you assume Figure 2-17 also applies to mice and you irradiate male sperm
with X rays (known to inactivate genes), what phenotype would you look for in
progeny in order to find cases of individuals with an inactivated SRY gene?
Answer: If we inactivate the SRY gene in mammals with radiation, the offspring
should all be phenotypically females, yet on the chromosome level there would
be both XX and XY (in this case sterile, female looking males).
11. In Figure 2-19, 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.
3 : 1 ratio means that all females have red eyes (X+/–), while half the males have
red (X+/Y) and half white (XW/Y).
Careful sex determination when counting F2 offspring would point out to a sex
12. In Figure 2-21, 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: The answer would be:
2/3 2/3 1/4 = 1/9 or 0.11
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 dominant
phenotype they could only be A/A or A/a. Now, probability that two
heterozygotes have a recessive homozygote offspring is 1/4.
13. 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. The middle part of generation II marriage shows a typical
test cross (expected 1:1). Neither ratio in the pedigree could be confirmed
because of a small sample size in any given family, but allele segregation is
14. Could the pedigree in Figure 2-31 be explained as an autosomal dominant
Chapter Two 9
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,
daughters will receive it only, and second, if mother has a gene, both sons and
daughters would receive it.
15. 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.
16. 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. Similarly, the pea has seven
different chromosomes and will produce seven bands (homologous
chromosomes will co-migrate as a single band).
17. The broad bean (Vicia faba) is diploid and 2n = 18. Each haploid chromosome
set contains approximately 4 m of DNA. The average size of each chromosome
during metaphase of mitosis is 13 m. 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 μm, 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.
18. 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
10 Chapter Two
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:
x+ haploid cells
2x diploid cells in G1 or cells after meiosis I but prior to meiosis II
4x diploid cells after S but prior to cell division
19. Name the key function of mitosis.
Answer: The key function of mitosis is to generate two daughter cells
genetically identical to the original parent cell.
20. 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.
21. Can you design a different nuclear-division system that would achieve the same
outcome as that of meiosis?
Answer: It’s pretty hard to beat several billions of 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.
22. 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- and long-term effects in such a
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. Parthenogenesis 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
Chapter Two 11
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 will be genetically identical to their mothers, and males would be
extinct within one generation.
23. 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 appears 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
24. 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 letters, 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 Distractions.
25. In an attempt to simplify meiosis for the benefit of students, mad scientists
develop a way of preventing 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
Answer: Yes, it could work but certain DNA repair mechanisms (such as
postreplication recombination repair) could not be invoked prior to cell division.
There would be just two cells as products of this meiosis, rather than four.
26. Theodor Boveri said, “The nucleus doesn’t divide; it is divided.” What was he
12 Chapter Two
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.
27. 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
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.
28. 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 (for example, the one
they inherited from their father or the one from their mother?), there are 223 =
8,388,608 possible combinations!
29. 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
pine tree sporophyte
sporophyte (ascus or
frog somatic cells gonads
butterfly somatic cells gonads
Chapter Two 13
snail somatic cells gonads
30. 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.
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
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
31. 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) chromosome movement to poles, (5) synapsis.
Answer: (5) chromosome pairing (synapsis)
32. In corn, the allele ƒ´ causes floury endosperm and the allele f´´ causes flinty
endosperm. In the cross ƒ´/ƒ´ [female symbol] ƒ´´/ƒ´´ [male symbol], 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
Answer: First, examine the crosses and the resulting genotypes of the
Female Male Polar
ƒ´/ƒ´ ƒ´´/ƒ´´ ƒ´ and ƒ´ ƒ´´/ƒ´´ ƒ´/ƒ´/ƒ´´
ƒ´´/ƒ´´ ƒ´/ƒ´ ƒ´´ and ƒ´´ ƒ´/ƒ´ ƒ´´/ƒ´´/ƒ´
14 Chapter Two
As can be seen, the phenotype of the endosperm correlates to the predominant
33. 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.
34. 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 test-cross (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
35. 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 phenotype? (Invent genetic symbols.)
b. What would an ascus from this cross look like?
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 normally sized 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 normally sized and 180 small-sized
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 two are s+.
Chapter Two 15
36. 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
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.
37. 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+ lys-5+
a. You expect two lys-5+ (black) spores and two lys-5 (white) spores.
b. You expect all lys-5 (white) spores.
c. You expect all lys-5+ (black) spores.
38. For a certain gene in a diploid organism, eight units of protein product are
needed for normal function. Each wild-type allele produces five units.
a. If a mutation creates a null allele, do you think this allele will be recessive
16 Chapter Two
b. What assumptions need to be made to answer part a?
a. You do not expect the mutation to be recessive. This would be an example
of a haploinsufficient gene since one copy of the wild-type allele does
produce enough protein product for normal function.
b. An important assumption would be that having five of eight units of protein
product would result in an observable phenotype. It also assumes that the
regulation of the single wild-type allele is not affected. Finally, if the mutant
allele was leaky rather than null, there might be sufficient protein function
when heterozygous with a wild-type allele.
39. A Neurospora colony at the edge of a plate seemed to be sparse (low density) in
comparison with the other colonies on the plate. This colony was thought to be
a possible mutant, and so it was removed and crossed with a wild type of the
opposite mating type. From this cross, 100 ascospore progeny were obtained.
None of the colonies from these ascospores was sparse, all appearing to be
normal. What is the simplest explanation of this result? How would you test
your explanation? (Note: Neurospora is haploid.)
Answer: The simplest explanation is that the abnormal phenotype was not due
to an genetic change. Perhaps the environment (edge of plate) was less
favorable for growth. Since Neurospora is haploid and forms ascospores,
isolating individual asci from a cross of the possible “mutant” to wild type and
individually growing the spores should yield 50 percent wild-type and 50
percent “mutant” colonies. If all spores yield wild-type colonies, the low density
phenotype was not heritable.
40. From a large-scale screen of many plants of Collinsia grandiflora, a plant with
three cotyledons was discovered (normally, there are two cotyledons). This
plant was crossed with a normal pure-breeding wild-type plant, and 600 seeds
from this cross were planted. There were 298 plants with two cotyledons and
302 with three cotyledons. What can be deduced about the inheritance of three
cotyledons? Invent gene symbols as part of your explanation.
Answer: Since half of the F1 progeny are mutant, it suggests that the mutation
that results in three cotyledons is dominant, and the original mutant was
heterozygous. Assuming C = the mutant allele and c = the wild-type allele, the
P C/c c/c
F1 C/c three cotyledons
c/c two cotyledons
41. In the plant Arabidopsis thaliana, a geneticist is interested in the development
Chapter Two 17
of trichomes (small projections). A large screen turns up two mutant plants (A
and B) that have no trichomes, and these mutants seem to be potentially useful
in studying trichome development. (If they were determined by single-gene
mutations, then finding the normal and abnormal functions of these genes
would be instructive.) Each plant is crossed with wild type; in both cases, the
next generation (F1) had normal trichomes. When F1 plants were selfed, the
resulting F2’s were as follows:
F2 from mutant A: 602 normal; 198 no trichomes
F2 from mutant B: 267 normal; 93 no trichomes
a. What do these results show? Include proposed genotypes of all plants in
b. Under your explanation to part a, is it possible to confidently predict the F1
from crossing the original mutant A with the original mutant B?
a. The data for both crosses suggest that both A and B mutant plants are
homozygous for recessive alleles. Both F2 crosses give 3:1 ratios of normal
to mutant progeny. For example, let A = normal and a = mutant, then
P A / A a/a
F2 1 A/A phenotype: normal
2 A/a phenotype: normal
1 a a phenotype: mutant (no trichomes).
b. No. You do not know if the a and b mutations are in the same or different
genes. If they are in the same gene then the F1 will all be mutant. If they are
in different genes, then the F1 will all be wild type.
42. You have three dice: one red (R), one green (G), and one blue (B). When all
three dice are rolled at the same time, calculate the probability of the following
a. 6 (R), 6 (G), 6 (B)
b. 6 (R), 5 (G), 6 (B)
c. 6 (R), 5 (G), 4 (B)
d. No sixes at all
e. A different number on all dice
Answer: Each die has six sides, so the probability of any one side (number) is
1/6. To get specific red, green, and blue numbers involves “and” statements that
are independent. So each independent probability is multiplied together.
a. (1/6)(1/6)(1/6) = (1/6)3 = 1/216
b. (1/6)(1/6)(1/6) = (1/6)3 = 1/216
c. (1/6)(1/6)(1/6) = (1/6)3 = 1/216
d. To not roll any sixes is the same as getting anything but sixes:
18 Chapter Two
(1 – 1/6)(1 – 1/6)(1 – 1/6) = (5/6)3 = 125/216.
e. The easiest way to approach this problem is to consider each die separately.
The first die thrown can be any number. Therefore, the probability for it is
The second die can be any number except the number obtained on the first
die. Therefore, the probability of not duplicating the first die is 1 – p(first
die duplicated) = 1 – 1/6 = 5/6.
The third die can be any number except the numbers obtained on the first
two dice. Therefore, the probability is 1 – p(first two dice duplicated) = 1 –
2/6 = 2/3.
Finally, the probability of all different dice is (1)(5/6)(2/3) = 10/18 = 5/9.
43. In the pedigree below, the black symbols represent individuals with a very rare
If you had no other information to go on, would you think it more likely that the
disease was dominant or recessive? Give your reasons.
Answer: You are told that the disease being followed in this pedigree is very
rare. If the allele that results in this disease is recessive, then the father would
have to be homozygous and the mother would have to be heterozygous for this
allele. On the other hand, if the trait is dominant, then all that is necessary to
explain the pedigree is that the father is heterozygous for the allele that causes
the disease. This is the better choice as it is more likely, given the rarity of the
44. a. The ability to taste the chemical phenylthiocarbamide is an autosomal
dominant phenotype, and the inability to taste it is recessive. If a taster
woman with a nontaster father marries a taster man who in a previous
marriage had a nontaster daughter, what is the probability that their first
child will be:
(1) A nontaster girl
(2) A taster girl
(3) A taster boy
b. What is the probability that their first two children will be tasters of either
a. By considering the pedigree (see below), you will discover that the cross in
question is T/t T/t. Therefore, the probability of being a taster is 3/4, and
the probability of being a nontaster is 1/4.
Chapter Two 19
Also, the probability of having a boy equals the probability of having a girl
(1) p(nontaster girl) = p(nontaster) p(girl) = 1/4 1/2 = 1/8
(2) p(taster girl) = p(taster) p(girl) = 3/4 1/2 = 3/8
(3) p(taster boy) = p(taster) p(boy) = 3/4 1/2 = 3/8
b. p(taster for first two children) = p(taster for first child) p(taster for second
child) = 3/4 3/4 = 9/16
45. John and Martha are contemplating having children, but John’s brother has
galactosemia (an autosomal recessive disease) and Martha’s great-grandmother
also had galactosemia. Martha has a sister who has three children, none of
whom have galactosemia. What is the probability that John and Martha’s first
child will have galactosemia?
Unpacking the Problem
1. Can the problem be restated as a pedigree? If so, write one.
Answer: Yes. The pedigree is given below.
2. Can parts of the problem be restated by using Punnett squares?
Answer: In order to state this problem as a Punnett square, you must first
know the genotypes of John and Martha. The genotypes can be
determined only through considering the pedigree. Even with the
20 Chapter Two
pedigree, however, the genotypes can be stated only as G/– for both John
The probability that John is carrying the allele for galactosemia is 2/3,
rather than the 1/2 that you might guess. To understand this, recall that
John’s parents must be heterozygous in order to have a child with the
recessive disorder while still being normal themselves (the assumption of
normalcy is based on the information given in the problem). John’s
parents were both G/g. A Punnett square for their mating would be:
The cross is:
P G/g G/g
g/g John’s brother
G/– John (either G/G or G/g)
The expected ratio of the F1 is 1 G/G : 2 G/g : 1 g/g. Because John does
not have galactosemia (an assumption based on the information given in
the problem), he can be either G/G or G/g, which occurs at a ratio of 1:2.
Therefore, his probability of carrying the g allele is 2/3.
The probability that Martha is carrying the g allele is based on the
following chain of logic. Her great-grandmother had galactosemia, which
means that she had to pass the allele to Martha’s grandparent. Because the
problem states nothing with regard to the grandparent’s phenotype, it must
be assumed that the grandparent was normal, or G/g. The probability that
the grandparent passed it to Martha’s parent is 1/2. Next, the probability
that Martha’s parent passed the allele to Martha is also 1/2, assuming that
the parent actually has it. Therefore, the probability that Martha’s parent
has the allele and passed it to Martha is 1/2 1/2, or 1/4.
John p(G/G) = 1/3
p(G/g) = 2/3
Martha p(G/G) = 3/4
p(G/g) = 1/4
This information does not fit easily into a Punnett square.
Chapter Two 21
3. Can parts of the problem be restated by using branch diagrams?
Answer: While the above information could be put into a branch diagram,
it does not easily fit into one and overcomplicates the problem, just as a
Punnett square would.
4. In the pedigree, identify a mating that illustrates Mendel’s first law.
Answer: The marriage between John’s parents illustrates Mendel’s first
5. Define all the scientific terms in the problem, and look up any other terms
about which you are uncertain.
Answer: The scientific words in this problem are galactosemia,
autosomal, and recessive.
Galactosemia is a metabolic disorder characterized by the absence of the
enzyme galactose-1-phosphate uridyl transferase, which results in an
accumulation of galactose. In the vast majority of cases, galactosemia
results in an enlarged liver, jaundice, vomiting, anorexia, lethargy, and
very early death if galactose is not omitted from the diet (initially, the
child obtains galactose from milk).
Autosomal refers to genes that are on the autosomes.
Recessive means that in order for an allele to be expressed, it must be the
only form of the gene present in the organism.
6. What assumptions need to be made in answering this problem?
Answer: The major assumption is that if nothing is stated about a person’s
phenotype, the person is of normal phenotype. Another assumption that
may be of value, but is not actually needed, is that all people marrying into
these two families are normal and do not carry the allele for galactosemia.
7. Which unmentioned family members must be considered? Why?
Answer: The people not mentioned in the problem, but who must be
considered, are John’s parents and Martha’s grandparent and parent
descended from her affected great-grandmother.
22 Chapter Two
8. What statistical rules might be relevant, and in what situations can they be
applied? Do such situations exist in this problem?
Answer: The major statistical rule needed to solve the problem is the
product rule (the “and” rule). It is used to calculate the cumulative
probabilities described in part 2 of this unpacked solution (e.g., What is
the probability that Martha’s parent inherited the galactosemia allele AND
passed that allele onto Martha AND Martha will pass that allele on to her
9. What are two generalities about autosomal recessive diseases in human
Answer: Autosomal recessive disorders are assumed to be rare and to
occur equally frequently in males and females. They are also assumed to
be expressed if the person is homozygous for the recessive genotype.
10. What is the relevance of the rareness of the phenotype under study in
pedigree analysis generally, and what can be inferred in this problem?
Answer: Rareness leads to the assumption that people who marry into a
family that is being studied do not carry the allele, which was assumed in
entry (6) above.
11. In this family, whose genotypes are certain and whose are uncertain?
Answer: The only certain genotypes in the pedigree are John’s parents,
John’s brother, and Martha’s great-grandmother and grandmother. All
other individuals have uncertain genotypes.
12. In what way is John’s side of the pedigree different from Martha’s side?
How does this difference affect your calculations?
Answer: John’s family can be treated simply as a heterozygous-byheterozygous
cross, with John having a 2/3 probability of being a carrier,
while it is unknown if either of Martha’s parents carry the allele.
Therefore Martha’s chance of being a carrier must be calculated as a series
13. Is there any irrelevant information in the problem as stated?
Chapter Two 23
Answer: The information regarding Martha’s sister and her children turns
out to be irrelevant to the problem.
14. In what way is solving this kind of problem similar to solving problems
that you have already successfully solved? In what way is it different?
Answer: The problem contains a number of assumptions that have not
been necessary in problem solving until now.
15. Can you make up a short story based on the human dilemma in this
Answer: Many scenarios are possible in response to this question.
Now try to solve the problem. If you are unable to do so, try to identify the
obstacle and write a sentence or two describing your difficulty. Then go back to
the expansion questions and see if any of them relate to your diffi
[Solved] TEST BANK FOR Introduction to Genetic Analysis, 10th Edition By Anthony J.F. Griffiths
- This solution is not purchased yet.
- Submitted On 15 Feb, 2022 02:57:19
- Rating : 0
- Grade : No Rating
- Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00