Genetics: A Conceptual Approach, 6th Edition Solution Manual

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
Chapter 1
Think-Pair-Share questions for the chapter opening story:
 Albinism occupied a special place in the Hopi culture; individuals who possessed this
trait were valued by members of the tribe. What are some examples of genetic traits that,
in contrast, sometimes result in discrimination and prejudice?
Possible Answers: People with many genetic traits are exposed to discrimination and
prejudice. Indeed, people with albinism in cultures other than the Hopis are subjected to
discrimination for the way they look. In the past, African Americans who were
heterozygous carriers of the mutation for sickle-cell anemia were subject to
discrimination in employment and insurance, even though they were perfectly healthy.
Skin color is a genetically determined characteristic and there is discrimination against
people with certain skin colors in some societies. Students may have witnessed
discrimination against people with a variety of genetically determined characteristics.
 Albinism in humans can be caused by mutations in any one of several different genes.
This situation, in which the same phenotype may result from variation in several different
genes, is referred to as genetic heterogeneity. Is genetic heterogeneity common? Are most
genetic traits in humans the result of variation in a single gene, or are there many genetic
traits that result from variation in several genes, as albinism does?
Possible Answers: Beginning genetics students often have the misconception that most
traits are caused by a single variation at a single gene, much like the traits Mendel studied
in his pea plants. This question provides an opportunity at the beginning of the course to
discuss this misconception and the fact that the genetic basis of most traits is more
complex.
Many genetic traits and diseases exhibit genetic heterogeneity, where DNA sequences at
any one of several different genes can result in the same phenotype. For example,
albinism in the Hopis is caused by mutations in the OCA2 gene, but mutations at several
other genes may produce albinism in other populations. Genetic heterogeneity can also
result from different mutations at the same gene. Both types of heterogeneity are seen in

Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
human albinism. Many genetic traits are also the result of combined variation at several
genes, a situation referred to as a polygenic trait (see Chapters 5 and 24).
End-of-chapter Think-Pair-Share questions:
Section 1.1
1. Bob says that he is healthy and has no genetic diseases such as hemophilia or Down
syndrome. Therefore, genetics, he says, plays little role in his life. Do you think Bob is
correct in his conclusion? Why or why not?
Possible Answers: Bob makes a common mistake when he thinks of genetic effects only
as classical genetic diseases, like hemophilia and Down syndrome. In fact, there are
many common disorders and diseases, as well as variations in healthy phenotypes, that
are influenced by genetics. In many of these cases, the phenotype is the result of a
variation in multiple genes (the trait is polygenic) and the interaction of genes and
environment (see Chapter 5). For example, common diseases, such as asthma, coronary
heart disease, hypertension, depression, and diabetes, are due to a combination of
variations at multiple genes and environmental factors. Many aspects of our appearance,
such as height, weight, facial features, hair color, skin color, and eye color, are also
determined at least to some degree by genetic variation.
2. Are mutations good or bad? Explain your answer.
Possible Answers: This question will likely elicit interesting responses from students.
Mutations might be seen as both good and bad. On the one hand, mutations provide the
necessary genetic variation necessary for evolution, the process by which populations
change and become adapted to their environments. Adaptation and evolutionary change
allow populations to avoid extinction when environments change. So, mutations can be
beneficial to the long-term success of a population. Mutations result in individual
differences, which makes each of us unique. Also, mutations are often useful in genetic
research, as we will see in our journey through the field of genetics. On the other hand,
when mutations result in genetic diseases, they may cause pain and suffering. Thus, many

Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
mutations also have detrimental effects.
Section 1.2
3. Do you support or oppose the development of genetically engineered foods (genetically
modified organisms, or GMOs)? Find someone who takes the opposite position and
discuss this question with them. Think about the economic, environmental, health, and
social effects of their use. List some reasons for and against genetically engineering the
foods we eat
Possible Answers: There are many possible answers to this question.
Some reasons for supporting the genetic engineering of foods include:
 Ability to provide more food to feed the world’s growing population
 Lowered use of pesticides and other chemicals that have negative health and
ecological effects
 Ability to produce foods at lower cost
 Ability to create crops that can grow in otherwise unsuitable habitats, such as arid
regions and saline soils
Some reasons for opposing genetically engineered foods include:
 Unanticipated health risks
 Potential ecological damage caused by release of novel genetically modified
organisms into the environment.
 Negative economic effects on small farmers who cannot afford the cost of
genetically engineered crops and animals.
Section 1.3
4. Why do you think all organisms use nucleic acids for encoding genetic information? Why
not use proteins or carbohydrates? What advantages might DNA have as the source of
genetic information?

Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
Possible Answers: One potential answer is that all organisms use nucleic acids because
all living organisms are derived from a common ancestor, which just happened to use
nucleic acids as its genetic information. However, there may also be reasons why nucleic
acids are advantageous as information carriers. Nucleic acids have the capacity to vary in
their base sequence, but they don’t vary too much. (There are only four common bases
used in DNA and RNA.) They have the ability to form long polymers, which means they
can encode a great deal of genetic information. DNA is particularly stable because of its
double-stranded nature and lack of any free hydroxyl groups. The double-stranded nature
of DNA also provides a particularly efficient means of replicating the molecule: the
complementary nature of the bases (G and C pair; A and T pair) means that each strand
of DNA can serve as a template for the synthesis of a new strand.
Chapter 2
Think-Pair-Share questions for the chapter opening story:
 In the blind men’s riddle, two blind men must sort out 10 pairs of socks so that each man
gets exactly five pairs of different colored socks. In the analogy, is it important that the
men are blind? In a cell, what does the blindness represent?
Possible Answers: The analogy depends on the men being blind. If they could see they
could distinguish the socks by color and easily sort the socks so that each man got two
socks of each color. While theoretically it might be possible for cells to evolve some
mechanism by which they distinguish each pair of chromosomes and by this
discrimination ensure that one chromosome of each pair ends up in a resulting cell, cells
have not evolved such a mechanism. Through the process of mitosis, each chromosome
replicates and the two replicated copies (analogous to the two socks of a pair) get pulled
in opposite directions and end up in separate cells. The cells don’t have a mechanism to
see the different types of chromosomes and ensure at the end that each daughter cell gets
exactly one copy of each pair—this is what the blindness represents.

Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
End-of-chapter Think-Pair-Share questions:
Section 2.2
1. A chromosome consists of two sister chromatids. Does the genetic information on the
two sister chromatids come from only one parent or from both parents? Explain your
reasoning.
Possible Answers: This questions is designed to help address the difference between
sister chromatids and homologous chromosomes, a common point of confusion for many
students. For the most part, the genetic information on the two sister chromosomes comes
from the same parent, because the sister chromatids are produced by replication and are
exact copies of the genetic information originally present on a single chromosome, which
comes from one parent. However, crossing over can produce information from two
different parents on the same chromosome. One chromosome of a homologous pair
comes from one parent and the other homolog comes from the other parent. When
crossing over takes place, information is exchanged between nonsister chromatids—
chromatids from different but homologous chromosomes. So, after crossing over, the
information on one chromatid may contain information from the other homolog (which
comes from the other parent).
2. Are homologous pairs of chromosome present in mitosis? Explain your reasoning.
Possible Answers: This question addresses a common misconception among students
who often assume that homologous pairs of chromosomes exist in meiosis but not in
mitosis. Homologous pairs of chromosomes are present in mitosis. However, they don’t
pair up in mitosis like they do in meiosis, and sister chromatids of each chromosome
separate independently.
3. A cell has eight chromosomes in metaphase II of meiosis. How many chromosomes and
DNA molecules will be present per cell in this same organism at the following stages?
a. Prophase of mitosis
b. Metaphase I of meiosis

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
c. Anaphase of mitosis
c. Anaphase II of meiosis
d. Anaphase I of meiosis
e. After cytokinesis that follows mitosis
f. After cytokinesis that follows meiosis II.
Possible Answers: This question requires that students think through the different steps
of mitosis and meiosis and fully understand what is happening in each stage.
Number of Number of
Chromosomes DNA Molecules
a. Prophase of mitosis 16 32
b. Metaphase I of meiosis 16 32
c. Anaphase of mitosis 32 32
c. Anaphase II of meiosis 16 16
d. Anaphase I of meiosis 16 32
e. After cytokinesis that follows mitosis 16 16
f. After cytokinesis that follows meiosis II 8 8
Section 2.3
4. What is the difference between sister chromatids and homologous chromosomes?
Possible Answers: Students often are confused and have misconceptions about the difference
between sister chromatids and homologous chromosomes. Sister chromatids are identical
copies (unless crossing over takes place) of the same original chromosome. Homologous
chromosomes are different chromosomes, containing information for the same traits
(homologous information) but not the same genetic information. One homolog comes from
one parent; the other homolog comes from the other parent.
5. List as many similarities and differences in mitosis and meiosis as you can. Which differences

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
do you think are most important and why?
Possible Answers:
Similarities
 Both involve chromosome and cell division.
 Both are preceded by DNA replication.
 Both use spindle fibers to separate chromosomes.
 Both have a stage where sister chromatids separate.
Differences
 Meiosis normally involves two cell divisions; mitosis usually has only a single cell
division.
 Chromosome reduction occurs in meiosis but not in mitosis.
 Resulting daughter cells are genetically different in meiosis but not in mitosis.
 Crossing over occurs in meiosis but does not normally take place in mitosis.
 Random assortment of chromosomes occurs in anaphase I of meiosis but does not occur
in anaphase of mitosis.
 In metaphase I of meiosis, pairs of homologous chromosomes line up; in metaphase of
mitosis (and metaphase II of meiosis) individual chromosomes line up.
 In anaphase I of meiosis, homologous chromosomes separate; in anaphase of mitosis (and
anaphase II of meiosis) sister chromatids separate.
6. Describe how and where each of the following terms applies to mitosis and/or meiosis: (1)
replication; (2) pairing; and (3) separation.
Possible Answers:
Replication, pairing, and separation are key events that take place in mitosis and meiosis.
Replication: In mitosis and meiosis, DNA replication takes place during S phase preceding
nuclear division.
Pairing: Homologous chromosomes pair up in meiosis, but no pairing of homologous
chromosomes takes place in mitosis. However, the sister chromatids of each chromosome are

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
paired in both mitosis and meiosis.
Separation: In mitosis, sister chromatids separate in anaphase. In meiosis, homologous
chromosomes separate in anaphase I and sister chromatids separate in anaphase II.
7. Do you know of any genetic diseases or disorders that result from errors in mitosis or
meiosis? How do errors in mitosis or meiosis bring about these diseases?
Possible Answers: There are many chromosome abnormalities that result from errors in
mitosis and/or meiosis, including Down syndrome (resulting from an extra copy of
chromosome 21) and Turner syndrome (resulting from a single X chromosome). Many
cancers have abnormal chromosomes that result from errors in mitosis (see Chapter 23).
Errors in mitosis and/or meiosis often result in abnormal separation of chromosomes, so that
cells end up with too many or too few chromosomes.
Chapter 3
Think-Pair-Share questions for the chapter opening story:
 Why is knowing about the genetic basis of a trait like blond hair important? Why would
scientists go to the trouble to investigate the genetic basis of blond hair in the Solomon
Islanders?
Possible Answers: There are a number of reasons why scientists might be interested in the
genetic basis of blond hair in Solomon Islanders. First is scientific curiosity. Many scientific
studies are conducted for no obviously practical reason, but simply for the purpose of
learning the answer to an interesting question. Another reason is that science advances by
gathering information about basic processes, in what is called basic science. It is often
difficult to know how scientific information might be useful in the future and so much of
science focuses on basic questions that have no current practical application. Often the
information gathered in basic research provides practical applications in the future.

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
There may also be practical reasons for studying the genetic basis of blond hair. One of the
hypotheses explaining blond hair in Solomon Islanders is that the population has some
European genes, so the study of this trait provides insight into the history and evolution of
Solomon Islanders. Blond hair results from reduced amounts of the melanin in hair.
Melanin plays a vital role in many processes, including vision and protection from UV
radiation (see the introduction to Chapter 1 about albinism). So the underlying genetics of
blond hair may lead to a better understanding of the biosynthesis of melanin and disorders
caused by reduced pigment.
 If a blond-haired person from northern Europe mated with a blond Solomon Islander,
what proportion of their offspring would be expected to have blond hair? Explain your
reasoning.
Possible Answer: Most likely, none of the offspring would have blond hair, because
different genes are coding for blond hair in Europeans and Solomon Islanders. The
European’s genotype might be aa B+B+, where aa codes for blond hair. The Solomon
islander might be A+A+ bb, where bb codes for blond hair. A mating between these two
individuals would be: aa B+B+ × A+A+ bb  A+a B+b, which would have nonblond hair.
End-of-chapter Think-Pair-Share questions:
Introduction
1. About 40% of Solomon Islanders carry a gene for blond hair, and yet only 5%–10% of these
people actually have blond hair. Why is the number of people with blond hair only 5%–10%
when so many people carry genes for blond hair?
Answer: This question focuses on the difference between genotypic and phenotypic
frequencies and how, with recessive traits, people can carry a gene for the trait (be
heterozygous) and yet not have the trait.
The allele for blond hair (b) is recessive to the allele for dark hair. Most of the people
who carry a blond allele are heterozygotes (Bb); because the blond allele is recessive, these
individuals have dark hair even though they carry a blond allele. If mating for hair color is

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
random, then the percent of people with blond hair is expected to be q2, where q equals the
frequency of the blond gene. If the percent of blond people is 10%, then q2 = 0.10. The
square root of 0.10 is 10.0 = 0.32 = q, the frequency of the blond gene. The frequency of
the dark-haired gene, p, will be 1 – q = 0.68. According to the Hardy-Weinberg law, which
we will learn in Chapter 25, the probability of a person being heterozygous is expected to be
2pq = 2(0.68)(0.32) = 0.44, which is about the frequency of dark-haired people on the
Solomon Islands who carry a blond gene.
2. Why was Mendel’s success dependent on his studying characteristics that exhibit only two
easily distinguished phenotypes, such as white versus gray seed coats and round versus
wrinkled seeds? Would he have been less successful if he had instead studied traits like seed
weight or length of the leaves, which vary much more in their phenotypes? Explain your
answer.
Possible Answer: This question is useful for discussing some of the factors that led to
Mendel’s success and the differences between discontinuous and continuous traits
(discussed in more detail in Chapter 24). Probably Mendel would have been less successful
if he had studied characteristics that vary more in phenotypes, such as seed weight or length
of the leaves. Because his characteristics differed in only two easily distinguishable
phenotypes, he was able to infer their genotypes. Many traits that vary continuously in
phenotype are encoded by multiple loci and often the phenotypes of these traits are
influenced by environmental effects (see Chapter 24).
Section 3.2
3. Geneticists often carry out reciprocal crosses when they are studying the inheritance of
traits. Why do geneticists use reciprocal crosses?
Possible Answer: Reciprocal crosses provide information about whether the sex of a
parent in a cross affects the types and proportions of progeny from the cross. For a simple
autosomal trait, reciprocal crosses usually give the same result. However, reciprocal
crosses do not give the same results if sex interacts with heredity, as it does for sex-linked,

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
sex-limited, sex-influenced, cytoplasmically inherited traits, genetic maternal effects, and
genomically imprinted traits (see Chapters 4 and 5). When these factors are present, the
inheritance of a trait may differ from that of simple autosomal traits, so it is important to
know whether these factors are present.
4. Red hair in humans is inherited as a recessive trait. Bill and Sarah both have black hair.
They marry and have four children, three of whom have red hair. Bill says it isn’t
genetically possible for two black-haired people to have ¾ red-haired children, and he
accuses Sarah of infidelity. Sarah says Bill is a homozygous dominant idiot and knows
nothing about genetics. Who is correct and why?
Possible Answer: This question helps to emphasize that the ratio of progeny in a genetic
cross often differs by chance from the expected ratio, especially when the number of
progeny is small.
Assume black hair is determined by a dominant allele R+ and red hair by a recessive allele
r. If Bill and Sarah both have black hair and have one or more red-haired children, then Bill
and Sarah are probably both heterozygous for red hair (R+r). Mating between Bill and Sarah
would produce: R+r × R+r
 ¾ R+_ (black hair) and ¼ rr (red hair), so the expectation is
that only one out of four children would have red hair. However, just by chance, they might
have three red-haired children. The probability of this can be determined by the expansion
of the binomial: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4, where a = the probability of black
hair (¾) and b = the probability of red hair (¼). In this case, the term 4ab3 equals the
probability of having four children, one with black hair and three with red hair, or 4(¾)(¼)3,
which equals 3/64 or 0.047. Thus, there is only about a 5% chance (one in 20) of getting
three children out of four with red hair when both parents are heterozygous. This is not very
likely but not impossible.
Section 3.3
5. Are Mendel’s principles of segregation and independent assortment even relevant today in the
age of genomics, where it is possible to sequence an organism’s entire DNA and determine all

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
of its genetic information? Why is it important to study these principles and how can they be
used?
Possible Answer: This question is useful for addressing a common misconception that
Mendel’s principles are outdated and no longer relevant because we can determine the entire
sequence of a genome. These principles are still relevant and important today because they
allow us to make predictions about the outcome of genetic crosses, which are of great
practical value in agriculture, biotechnology, and human heredity. Having the DNA sequence
information does not, by itself, provide us with information about individual genes, the
dominance relationships of alleles, or whether genes assort independently, characteristics that
are essential for making predictions about heredity.
6. In cats, short hair (L) is dominant to long hair (l) and stripes (A) are dominant to solid color
(a). A cat with genotype Ll Aa mates with a cat that is Ll aa and they produce a litter of six
kittens. What is the probability that four of six kittens will have both long hair and stripes?
Answer: The cross for hair length is: Ll × Ll  ¾ L_ (short) and ¼ ll (long). The cross for
stripes is Aa × aa  ½ Aa (stripes) and ½ aa (solid). Using the multiplication rule, the
probability of a kitten having long hair and strips (ll Aa) is: ¼ × ½ = 1/8. The probability of a
kitten not having long hair and stripes is 1 – 1/8 = 7/8. We can use the binomial expansion (a +
b)6 to determine the probability that in a litter of six kittens, four will have long hair and
stripes, where a = the probability of having long hair and stripes and b = the probability of not
having long hair and stripes. The binomial expansion is: a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4
+ 6ab5 + b6. The term 15a4b2 represents the probability of four kittens with long hair and
stripes in a litter of six. Substituting in the values of a and b, we obtain: 15(1/8)4(7/8)2 = 0.0028.
Section 3.4
7. In corn, purple kernels (P) are dominant to yellow kernels (p) and starchy kernels (Su) are
dominant to sugary kernels (su). A corn plant grown from a purple and starchy kernel is
crossed with a plant grown from a yellow and sugary kernel and the following progeny
(kernels) are produced:

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
Phenotype Number
purple starchy 150
purple, sugary 142
yellow, starchy 161
yellow, sugary 115
Formulate a hypothesis about the genotypes of the parents and offspring in this cross. Perform
a chi-square goodness-of-fit test comparing the observed progeny with the numbers expected
based on your genetic hypothesis. What conclusion can you draw based on the results of your
chi-square test? Can you suggest an explanation for the observed results?
Answer: The cross is purple and starchy kernels × yellow and sugary kernels. The presence of
yellow and sugary kernels in the progeny tells us that the purple and starchy parent is
heterozygous for both purple and starchy: Pp Susu × pp susu. This cross is expected to
produce a 1:1:1:1 progeny ratio:
Pp Susu × pp susu
¼ Pp Susu purple, starchy
¼ Pp susu purple, sugary
¼ pp Susu yellow, starchy
¼ pp susu yellow, sugary
The observed and expected progeny are:
Phenotype Observed Expected
purple starchy 150 ¼ × 568 = 142
purple, sugary 142 ¼ × 568 = 142
yellow, starchy 161 ¼ × 568 = 142
yellow, sugary 115 ¼ × 568 = 142
total 568

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
The chi-square value is  2
2 observed exp ected
exp ected
 
  . Substituting the observed and
expected values, we obtain: 12.8
142
)142115(
142
)142161(
142
)142142(
142
)142150( 2222








The degrees of freedom are n – 1 = 4 – 1 = 3
From Table 3.7, we find the probability associated with a chi-square value of 8.12 and 3 d.f.
is less than 0.05. Therefore, the probability that the observed would deviate from the expected
by chance alone is low and we reject the hypothesis that this represents a 1:1:1:1 ratio. All of
the observed numbers are close to those expected with the exception of the yellow and sugary
phenotype. There are far fewer yellow sugary progeny than are expected. One possible
explanation is that the yellow sugary progeny exhibits lower probability of developing.
Perhaps some of them died at an early developmental stage and therefore do not appear
among the progeny.
Chapter 4
Think-Pair-Share questions for the chapter opening story:
 Climate warming could lead to all-female populations of bearded dragons, resulting in
extinction of the species. What are some other potential effects of climate warming on
natural populations of organisms?
Possible Answers: There are many possible answers to this question. For many organisms,
temperature plays an important role in many aspects of their ecology, and organisms are
frequently genetically adapted to function at a particular temperature. A change in
temperature may make them unable to survive and reproduce, leading to population declines
and eventual extinction. For example, breeding in many amphibians is triggered by early
spring temperatures. A rise in temperature may cause them to breed earlier, when food and
sufficient rainfall may not be available for the development of tadpoles. Many organisms
such as trout are adapted to cooler temperatures and may not be able to survive and
reproduce in water that has warmed because of climate change. Climate warming will bring

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e
less rainfall to some areas, leading to the local extinction of many populations of plants and
animals. Melting of ice may lead to sea-level rise, which may make it difficult for some
organisms to survive. Warmer temperatures may increase the prevalence of pathogens and
vectors of disease, leading to more disease. Some organisms may be able to genetically
adapt to these changes (see Chapters 25 and 26), but others may not and might go extinct.
 High incubation temperatures reverse the sex of ZZ bearded dragons, causing them to
develop as females instead of males. Imagine that temperature also caused sex reversal in
ZW individuals, so that high temperatures caused ZW individuals to develop as males
instead of females (this doesn’t actually happen in the bearded dragons). What would be
the result of a mating between a normal ZW female and a sex-reversed ZW male?
Answer: A cross between a ZW female and sex reversed ZW male would produce: ZW ×
ZW  ¼ ZZ, ½ ZW, ¼ WW. At 32o C, the ZZ individuals would develop as male and the
ZW would develop as female. The W chromosome is like the Y chromosome, with little
genetic information. A WW zygote, like a YY zygote, would probably be nonviable.
Therefore, the sex ratio of the offspring would become 1/3 ZZ males and 2/3 ZW females. At
higher temperatures, some ZW individuals might develop as males, skewing the sex ratio
towards more males.
End-of-chapter Think-Pair-Share questions:
Section 4.1
1. The duck-billed platypus has a unique mechanism of sex determination: females have five
pairs of X chromosomes (X1X1X2X2X3X3X4X4X5X5) and males have five pairs of X and Y
chromosomes (X1Y1X2Y2X3Y3X4Y4X5Y5). Do you think each of the X and Y chromosome
pairs in males assort independently of other X and Y pairs during meiosis? Why or why not?
Possible Answer: The individual pairs of X and Y chromosomes in male platypuses must not
separate independently. If they did separate independently, gametes with differing numbers of
X and Y chromosomes would result, which would produce offspring with abnormal numbers
of sex chromosomes. For example, independent separation of X and Ys in males would

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