Download Diagnostic Test Page 39 1. The correct answer is B. Based on

Chapter 11
Teacher Guide and Answers
Diagnostic Test
• Student thinks both A and B are recessive.
Direct student to discussion on multiple alleles
in Section 2.
• Student thinks blood types differ among people with an allele for both A and B blood types.
Direct student to the multiple alleles discussion
in Section 2.
Page 39
1. The correct answer is B. Based on student
responses, use the list below to address
preconceptions.
• Student thinks that there are no diseases or
disorders caused by heredity. Direct student
Launch Lab
to the recessive genetic disorders discussion in
Section 1.
Page 40 • What do you know about human
• Student thinks recessive genes cause all genetic inheritance?
disorders. Direct student to the dominant
Analysis
genetic disorders discussion in Section 1.
1. Answers will vary, but should give some
• Student thinks dominant genes cause all
insight into the knowledge, background, and
genetic disorders. Direct student to the recessive
experiences your students have regarding
genetic disorders discussion in Section 1.
human heredity. Identify and correct any
• Student thinks environmental factors cause
misconceptions.
albinism. Direct student to the recessive genetic
2. Knowledge of human heredity is necessary to
disorders discussion in Section 1.
understand legal, social, and moral issues that
• Student thinks dietary factors cause albinism.
involve inherited traits. Such knowledge could
Direct student to the recessive genetic disorders
help with making certain health decisions.
discussion in Section 1.
2. The correct answer is C. Based on student
responses, use the list below to address
preconceptions.
• Student thinks a pedigree catalogs all observable traits of a person. Direct student to the
pedigree discussion in Section 1.
• Student thinks a pedigree traces all the observable traits through a family’s history. Direct
student to the pedigree discussion in Section 1.
• Student thinks a pedigree only involves the
family history of royalty or famous people.
Direct student to the pedigree discussion in
Section 1.
• Student thinks a pedigree maps the genetic
makeup of a person. Direct student to the pedigree discussion in Section 1.
3. Neither A nor B is dominant. The two blood
types are codominant, which means that both
alleles are expressed. Based on student responses,
use the list below to address preconceptions.
• Student thinks either A or B is the dominant
allele. Explain to student that both blood types
are dominant because both alleles are expressed.
The alleles are codominant.
Unit 3
MiniLab
Page 41 • Investigate Human Pedigrees
Analysis
1. With pedigrees, it is easy to follow traits from
one generation to the next.
2. Families affected with unfavorable traits can be
given advice about the chances of their future
children possessing these traits. However,
pedigree information obtained from only a few
members of a family could be inaccurate, unreliable, or misleading.
Minilab
Page 42 • Explore the Methods of the Geneticist
Analysis
1. Answers will vary. Sample answer: We look for
the ratio of hitchhiter’s thumbs to
non-hitchhiter’s thumbs.
2. Students might suggest DNA analysis or compiling pedigrees to determine dominance. In small
populations, traits can be more common even
though they are recessive, which might cause
students to misidentify them as dominant.
CHAPTER 11 TEACHER GUIDE AND ANSWERS 155
Chapter 11
Teacher Guide and Answers
Part B
• Ask students “How is genetic engineering similar
Page 43 • What’s in a face? Investigate Inherited to selective breeding?” “In what ways is genetic
Human Facial Characteristics
engineering different from what has been practiced
Analyze and Conclude
as selective breeding?” Discuss the concept of unintended outcomes, and elicit some examples from
1. The male determines the gender of the offspring
students’ everyday experiences and observations.
in humans.
• Ask students to interview their parents or other
2. a 50 percent chance in each situation
older relatives or friends about what they remem3. To achieve this outcome, both parents must have
ber about thalidomide babies or DDT. Have stuwavy hair (Hh).
dents share their findings. Discuss the intended
4. Answers will depend on the traits used.
purposes of the substances and their unintended
5. The chances of two groups producing identical
outcomes. Extend these concepts to the informaoffspring are quite remote. It would require each
tion provided in the table in this activity.
coin flip for each group to be exactly the same
• Below Level: (1) Provide an audiotape of the text to
for each trait.
which students can listen while reading the activity.
(2) Laminated cutouts of the ears of corn and the
Real-World Biology: Analysis
corn plants in the different sizes and conditions of
“robustness” can be provided as manipulatives to
Page 45 • Improving Food Crops
aid in understanding the hybridization process.
Planning the Activity
• Above Level: Have students choose an issue or a
This activity may be used with the study of genetproblem from the table, plan a study of the probics to reinforce and extend the concepts of selective
lem, and explain how data will be collected and
breeding and genetic engineering.
analyzed.
Purpose
Answers to Student Worksheet
Students examine methods and effects of selective
Part A: Hybrid Vigor
breeding and genetic engineering.
BioLab
Career Applications
Many agricultural technicians work under the
guidance of agricultural scientists as they conduct
tests and experiments to improve the yield and
quality of crops or to increase the resistance of plants
and animals to disease, insects, or other hazards.
In some cases, technicians use computers and
computer-interfaced equipment extensively during
their research. In other cases, technicians perform
much of their work outdoors, sometimes in remote
locations.
Teaching Strategies
Part A
Introduce the concept of selective breeding by eliciting from students information about what they
know about pedigreed animals. What does pedigree
mean? Do students own a pedigreed dog or cat?
If so, what are the animal’s characteristics? Which
are the most desirable?
156 CHAPTER 11 TEACHER GUIDE AND ANSWERS
Analyze and Conclude
1. Two inbred strains, A and B, are crossed with
each other to produce a hybrid A × B plant. Two
other inbred strains, C and D, are crossed with
each other to produce a hybrid C × D plant.
Then plants A × B and C × D are crossed to produce a double hybrid (A × B) × (C × D) plant.
2. larger ears of corn and taller plants
Part B: Genetically Engineered Foods
Analyze and Conclude
1. Scenarios will vary, but should be based on current evidence and employ logical reasoning.
Careers in Biology
Agricultural technicians conduct research, tests,
and experiments to improve the yield and quality of
crops or to increase the resistance of plants and animals to diseases, insects, or other hazards.
Unit 3
Chapter 11
Teacher Guide and Answers
Enrichment
Page 47 • Blood Typing in Forensic Science
System
Groups
Frequency in the General Population (average)
ABO
A, B, AB, O
A
AB
=
=
40%
5%
B
O
=
=
10%
45%
Rh
Rh+, Rh–
Rh+
=
85%
Rh–
=
15%
MN
M, N
M
=
50%
N
=
50%
Lewis
Le(a+b–), Le(a–
b+), Le(a+b+),
Le(a–b–)
Le(a+b–)
Le(a+b+)
=
=
22%
6%
Le(a–b+)
Le(a–b–)
=
=
72%
~0%
Cartwright
Yta, Ytb
Yta
=
99%
Ytb
=
1%
Student calculations will differ depending on the
blood systems chosen and the blood groups assigned
to the individual for whom the calculation is
performed.
Concept Mapping
Page 48 • Genetic Disorders
1. a dominant gene
2. nondisjunction
3. albinism
4. Huntington’s disease
5. too many
6. Down syndrome
7. Turner’s syndrome
Study Guide
Page 49 • Section 1
1. pedigree
2. alleles
3. homozygous
4. alleles
5. heterozygous
6. dominant
7. recessive
8. cystic fibrosis or albinism
9. albinism or cystic fibrosis
Unit 3
Students will discover that the frequency of blood
groups might differ among different populations.
The numbers given in the above table are rough estimates for the American Caucasian population.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
albinism
galactosemia
Tay-Sachs disease
cystic fibrosis
Huntington’s disease
achondroplasia
true
false
false
true
true
recessive
They are carriers.
There has to be at least one dominant gene.
Page 51 • Section 2
1. The check should be in the far right cell.
2. The check should be in the middle cell.
3. The check should be in the far left cell.
4. no
5. no
6. male
CHAPTER 11 TEACHER GUIDE AND ANSWERS 157
Chapter 11
Teacher Guide and Answers
Page 52 • Section 3
1. A
2. D
3. E
4. B
5. C
6. There are two copies of chromosome 20.
7. There are three copies of chromosome 21.
8. There are two copies of chromosome 22.
9. It has a set of three chromosomes of one kind.
10. nondisjunction
11. older
Guía de estudio
Página 53 • Sección 1
1. pedigrí
2. alelos
3. homocigoso
4. alelos
5. heterócigo
6. dominante
7. recesivo
8. fibrosis quística o albinismo
9. albinismo o fibrosis quística
10. albinismo
11. galactosemia
12. enfermedad de Tay-Sachs
13. fibrosis quística
14. enfermedad de Huntington
15. acondroplasia
16. verdadero
17. falso
18. falso
19. verdadero
20. verdadero
21. recesivo
22. Son portadores.
23. Tiene que haber al menos un gen dominante.
158 CHAPTER 11 TEACHER GUIDE AND ANSWERS
Página 55 • Sección 2
1. Se debe marcar la celda de la extrema derecha.
2. Se debe marcar la celda del medio.
3. Se debe marcar la celda de la extrema izquierda.
4. no
5. no
6. hombre
Página 56 • Sección 3
1. A
2. D
3. E
4. B
5. C
6. Hay dos copias de cromosoma 20.
7. Hay tres copias de cromosoma 21.
8. Hay dos copias de cromosoma 22.
9. Tiene un conjunto de tres cromosomas de una
clase.
10. no disyunción
11. mayores
Section Quick Check
Page 57 • Section 1
1. Student answers will vary. Answers might
include three of the following recessive genetic
disorders: cystic fibrosis, albinism, Tay-Sachs
disease, galactosemia, and sickle-cell anemia.
2. A pedigree is a diagram that traces the inheritance of a particular trait through several
generations.
3. CC is homozygous dominant, Cc is a carrier, and
cc is heterozygous recessive. The individual with
the cc genotype will have cystic fibrosis.
4. Their offspring will not express the recessive
trait because both parents must be carriers (have
at least one recessive allele) to produce offspring
with two recessive alleles.
5. Huntington’s disease is not lethal until after
reproductive age, so individuals can pass on
the allele to their children before they have
symptoms.
Unit 3
Chapter 11
Teacher Guide and Answers
Page 58 • Section 2
1. Codominance happens when both alleles are
expressed in the heterozygous condition.
2. Coat color of rabbits is determined by multiple
alleles.
3. Certain phenotypes or disorders that are inherited can be affected by environment. Student
examples will vary, but might include that the
tendency to develop heart disease is inherited,
but the occurrence and seriousness of the disease
are affected by diet and exercise.
4. With multiple alleles, more than two alleles are
possible for one pair of genes. For polygenic
traits, multiple alleles interact in multiple pairs
of genes.
5. For a woman to be color blind, the mother must
be a carrier or homozygous color blind, and the
father must be color blind. Because color blindness is rare, it would be rare for two such people
to meet and have children.
Page 59 • Section 3
1. Down syndrome is usually caused by three copies of chromosome 21.
2. A telomere is a protective cap that is found on
the end of chromosomes and consists of DNA
associated with proteins.
3. A karyotype is a micrograph in which pairs
of homologous chromosomes are arranged in
decreasing size.
4. Genotype XXX contains three copies of the
X chromosome, so it is trisomy.
5. Amniocentesis samples the fluid surrounding
the fetus for various chemicals and tests for certain genetic disorders. Chorionic villus sampling
samples placenta tissue to test for chromosomal
abnormalities. Chorionic villus sampling can
be done earlier in the pregnancy but might not
reveal problems with metabolism and is less
accurate than amniocentesis.
Chapter Test A
Page 60 • Part A: Multiple Choice
1. C
2. D
3. B
Unit 3
Page 60 • Part B: Matching
1. B
2. C
3. A
Page 61 • Part C: Interpreting Pedigrees
1. The Roman numerals represent successive
generations.
2. one
3. The pedigree illustrates a dominant genetic
disorder.
4. The black circle could represent a female that
is either heterozygous or homozygous for the
disorder.
Page 62 • Part D: Short Answer
1. Many inherited traits involve two forms of
alleles for a trait, but human blood types have
three forms of alleles including O, A, and B.
Different combinations of these alleles create
several different blood types.
2. Sex-linked traits are controlled by genes
located on the X chromosome. Males have one
X chromosome, and females have two X chromosomes. As a result, males are affected by sexlinked traits more often than females.
Page 62 • Part E: Concept Application
1. Cystic fibrosis is a recessive genetic disorder.
For the outward expression of the disorder to
express itself, a person must have two alleles for
the disease. A carrier of the disorder will not
express the disease outwardly but will carry one
allele for the disorder.
2. The statement is incorrect. Down syndrome
cannot be cured because the disease is a result
of a mutation in the chromosomes of the individual. An extra chromosome 21 is the cause of
Down syndrome, and it is currently impossible
to remove that chromosome from every cell.
Chapter Test B
Page 63 • Part A: Multiple Choice
1. D
2. A
CHAPTER 11 TEACHER GUIDE AND ANSWERS 159
Chapter 11
Teacher Guide and Answers
3. B
4. B
Page 63 • Part B: Matching and Completion
Matching
1. E
2. F
3. B
4. A
5. C
Completion
6. carrier
7. incomplete dominance
8. epistasis
9. autosomes
10. polygenic traits
Page 64 • Part C: Interpreting Pedigrees
1. Both parents are carriers of the disease and
are heterozygous for Tay-Sachs disease.
2. The children would either be homozygous
for not having the gene or heterozygous carriers
of the disease.
3. II3 and II6
4. The first-generation female must be heterozygous for the disorder. If the female were
homozygous for the disorder, all her children
would express the disorder because it is a dominant genetic disorder.
Page 65 • Part E: Concept Application
1. By analyzing the pedigrees of both families,
the Tay-Sachs trait can be studied, and a genetic
counselor can try to infer the genotypes of each
set of parents from the observation of the phenotypes of family members. The genetic counselor can help the couple determine whether the
inheritance pattern for the disorder is dominant
or recessive, and the genotypes of the man and
woman can be determined. The couple can use
this information to decide whether or not they
want to have biological children.
2. The statement is incorrect. Red-green color
blindness is a recessive X-linked trait. Because
males have only one X chromosome, they are
more likely to be colorblind because only one
recessive allele for the trait will cause the disorder to be outwardly expressed. Females have
two X chromosomes, and a female must have
two recessive alleles for the trait to be outwardly
expressed. This condition is less likely, but it
does occur.
Chapter Test C
Page 66 • Part A: Multiple Choice
1. A
2. C
3. C
4. A
5. D
6. D
Page 65 • Part D: Short Answer
1. The genotypes are homozygous recessive (cc)
Page 66 • Part B: Completion
and heterozygous (Cc). The phenotype of a per1. carrier
son who is homozygous recessive is the presence
2. incomplete dominance
of all sickle-shaped red blood cells, while the
3. Epistasis
phenotype of a person who is heterozygous for
4. polygenic traits
the disorder will have both normal-shaped and
5. concordance rate
sickle-shaped red blood cells.
6. fetal blood sampling
2. Blood types are determined by the multiple
alleles A, B, and O. Blood type A has A markers, Page 67 • Part C: Interpreting Pedigrees
blood type B has B markers, and blood type O
1. II1: homozygous for not carrying the Tayis the absence of either A or B markers. When
Sachs disease gene; II2: heterozygous carrier of
the A and B markers are codominant, they form
the disease; II3: homozygous for not carrying
blood type AB.
the Tay-Sachs disease gene; II4: heterozygous
carrier of the disease.
160 CHAPTER 11 TEACHER GUIDE AND ANSWERS
Unit 3
Chapter 11
Teacher Guide and Answers
2. One-fourth of children will be homozygous
for the disease, ½ of the children will be heterozygous carriers of the disease, and ¼ will
be homozygous for not carrying the Tay-Sachs
disease gene. The phenotypes would be ¼ of the
children would outwardly express the gene and
¾ would not express the gene.
3. I2: heterozygous dominant; II2: heterozygous dominant; II3: homozygous recessive; II4:
heterozygous dominant
4. The genotypes would be: ¼ homozygous
dominant, ½ heterozygous dominant, and ¼
homozygous recessive. The phenotypes would be
¾ of the children would outwardly express the
gene, and ¼ would not express the gene.
Page 68 • Part D: Short Answer
1. By analyzing a family’s history, the trait of
a genetic disease can be studied, and a genetic
counselor can try to infer the genotypes of a
person’s parents from the observation of phenotypes of family members. The genetic counselor
can help the patient determine whether the
inheritance pattern for the disorder is dominant
or recessive, and the genotype of the individual
can be determined.
2. Both X and Y chromosomes are sex chromosomes. The X chromosome is larger and contains genes essential for the development of both
males and females. The Y chromosome primarily contains genes necessary for the development
of male characteristics.
Page 68 • Part E: Concept Application
1. The inheritance pattern is called codominance. All the offspring have an allele for red
hair and an allele for white hair, and the traits
of both alleles are expressed. In a codominant
inheritance pattern, there is no recessive allele.
2. Both parents could be homozygous or heterozygous for type A blood. Another possibility
is that one parent could be homozygous or heterozygous for type A blood, and the second parent could have type O blood.
Unit 3
CHAPTER 11 TEACHER GUIDE AND ANSWERS 161