Behavioral Genetics Sixth Edition Test Bank

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TEST BANK
to accompany
Robert Plomin
John C. DeFries
Valerie S. Knopik
Jenae M. Neiderhiser
BEHAVIORAL GENETICS
Sixth Edition
Christina Hewitt
University of Colorado Boulder
WORTH PUBLISHERS
Test Bank
by Christina Hewitt
to accompany

Robert Plomin, John C. Defries, Valerie S. Knopik and Jenae M. Neiderhiser
Behavioral Genetics, Sixth Edition
© 2013 by Worth Publishers
All rights reserved.
The contents, or parts thereof, may be reproduced for use with Behavioral Genetics,
Sixth Edition by Robert Plomin, John C. Defries, Valerie S. Knopik and Jenae M. Neiderhiser,
but may not be reproduced in any form for any other purposes without
prior written permission of the publisher.
ISBN-10: 1-4641-4199-1
ISBN-13: 978-1-4641-4199-7
First Printing
Worth Publishers
41 Madison Avenue
New York, NY 10010
www.worthpublishers.com
Contents
CHAPTER 1
Overview
CHAPTER 2
Mendel’s Laws of Heredity
CHAPTER 3
Beyond Mendel’s Laws
CHAPTER 4
DNA: The Basis of Heredity
CHAPTER 5
Animal Models in Behavioral Genetics

CHAPTER 6
Nature, Nurture, and Human Behavior
CHAPTER 7
Estimating Genetic and Environmental Influences
CHAPTER 8
The Interplay between Genes and Environment
CHAPTER 9
Identifying Genes
CHAPTER 10
Pathways between Genes and Behavior
CHAPTER 11
Cognitive Disabilities
CHAPTER 12
General Cognitive Ability
CHAPTER 13
Specific Cognitive Abilities
CHAPTER 14
Schizophrenia
CHAPTER 15
Other Adult Psychopathology
CHAPTER 16
Developmental Psychopathology
CHAPTER 17
Personality And Personality Disorders
CHAPTER 18
Substance Use Disorders
CHAPTER 19
Health Psychology and Aging
CHAPTER 20
Evolution and Behavior
CHAPTER 21
The Future of Behavioral Genetics

CHAPTER 1
Overview
VOCABULARY
Match the phrases to the terms below. Each term may be used once, more than once, or not at all.
Phrases
1.) A self-replicating, DNA containing structure found in the nucleus of a cell
2.) The basic unit of heredity
3.) One of the variant forms of a gene
4.) A test used to discover if a fetus carries a genetic abnormality
5.) The complete DNA sequence of an individual
6.) The molecule that encodes genetic information
7.) The proportion of total variation in phenotype that is due to genetic variation
Terms
allele
amniocentesis
chromosome
clone
correlation
DNA
fragile X
gene
genome
heritability
multivariate genetic analysis
Ans:
1.) chromosome 2.) gene 3.) allele 4.) amniocentesis 5.) genome 6.) DNA 7.) heritability
Level: M
Topic: Overview
Pages: 1–5 and Glossary
TRUE / FALSE
1. One of the most important changes in the behavioral sciences in the past 30 years is the recognition that genes
influence almost all behaviors
Ans: True
Level: E
Page: 1
Topic: Overview
2. If genes influence a behavior, we would expect identical twins to be more similar for that behavior than

non-identical (fraternal) twins.
Ans: True
Level: M
Page: 1
Topic: Overview
3. Among the childhood psychopathologies, autism and ADHD are the most genetically influenced.
Ans: True
Level: E
Page: 1
Topic: Overview
4. Antisocial behavior in adolescence is highly heritable.
Ans: False
Level: M
Page: 1
Topic: Overview
5. Genes only influence disordered behavior.
Ans: False
Level: E
Page: 1
Topic: Overview
6. Weight is almost as influenced by genes as height.
Ans: True
Level: E
Page: 2
Topic: Overview
7. Personality and cognitive ability vary from person to person in large part due to genetic differences between
people.
Ans: True
Level: D
Page: 2
Topic: Overview
8. Many cases of early-onset Alzheimer disease are caused by single genes.
Ans: True
Level: E
Page: 2
Topic: Overview
9. No cure for Alzheimer disease has been discovered, so it is worthless that single-gene causes for the disease
have been identified.
Ans: False
Level: D
Page: 2
Topic: Overview
10. Late-onset Alzheimer’s disease is not influenced by genetics.

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Ans: False
Level: M
Page: 2
Topic: Overview
11. It is possible to get a genetic test for a gene-variant that influences risk for late-onset Alzheimer disease.
Ans: True
Level: D
Page: 2
Topic: Overview
12. Down syndrome is caused by a problem in a single gene.
Ans: False
Level: M
Page: 2
Topic: Overview
13. Gene influence on cognitive ability decreases over the course of a lifetime.
Ans: False
Level: E
Page: 3
Topic: Overview
14. Cognitive ability in adulthood is not nearly as influenced by genes as is height.
Ans: False
Level: M
Page: 3
Topic: Overview
15. The family environment is the most important cause of similarities in behavior among family members.
Ans: False
Level: M
Page: 3
Topic: Overview
16. The environment is not an important cause of individual differences in behavior.
Ans: False
Level: M
Page: 3
Topic: Overview
PROBLEMS
1. Parents who play chess with their children have children who do better in school. Does this mean that playing
chess is an environmental cause for better academic achievement?
Ans: Not necessarily. A correlation does not prove causation. It might be that parents and children who like to
play chess together do so because they share similar genetic constitutions and this constitution makes the
children do better in school.
Level: M
Topic: overview

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Pages: 3, 5
2. Explain the statement, “To some extent, people create their own experiences for genetic reasons.”
Ans: Since many environmental measures used in behavioral sciences show genetic influences, it is likely that
these environmental outcomes (like parenting, childhood accidents, life events, social activities, etc.) come
about in part because of the genetic constitution of the person, acting through characteristics like personality
and cognitive style.
Level: D
Topic: overview
Pages: 3, 5

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CHAPTER 2
Mendel’s Laws of Heredity
VOCABULARY
Match the phrases to the terms below. Each term may be used once, more than once, or not at all.
Phrases
1.) An alternative form of a gene at a locus
2.) Describes an allele that needs two copies to be present to produce its phenotype
3.) An example of a disorder caused by a dominant allele
4.) Describes an allele that produces a particular phenotype even when only one copy is present
5.) The genetic constitution of an individual, or the combination of alleles at a particular locus
6.) An example of a disorder caused by a recessive allele
7.) Cell division during which crossing-over occurs
8.) Type of cell division that occurs in ovaries and testes, producing gametes
9.) A basic unit of inheritance, found on a chromosome
10.) Said to be present for genes close together on a chromosome
Terms
allele
amino acid
chromatid
chromosome
dominant
gamete
gene
genotype
Huntington disease
linkage
locus
meiosis
pedigree
phenotype
phenylalanine
phenylketonuria (PKU)
protein
recessive
recombination
Ans:
1.) allele 2.) recessive 3.) Huntington disease 4.) dominant 5.) genotype 6.) phenylketonuria 7.)
meiosis 8.) meiosis 9.) gene 10.) linkage
Level: E
Topic: Mendel’s Laws of Inheritance

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Pages: 6–19
PROBLEMS
1. The albino phenotype is an inherited trait that has behavioral consequences in mice. In the offspring of two
carrier mice, the Mendelian phenotypic ratio of 3 normal to 1 albino is observed. How many genes does this
suggest albinism is controlled by?
Ans: one
Level: M
Topic: Mendel’s First Law of Heredity
Pages: 8–10
2. The behavioral Waltzer phenotype in mice is characterized by head shaking, rapid circling, and irritability
(van Abeelen and van der Kroon, 1967). True-breeding mice are mice that have two similar alleles at the
Waltzer locus.
True-breeding Waltzer males and females, when mated, produce offspring all with the Waltzer phenotype.
True-breeding Waltzer females and non-Waltzer males produce all normal offspring. Second generation (F2)
results from mating the offspring from a cross between true-breeding Waltzer and non-Waltzer mice are
shown below:
Normal (non-Waltzer) mice 124
Waltzer mice 48
TOTAL 172
a. Do you think the allele for the Waltzer phenotype is dominant or recessive?
Ans: recessive
Level: E
Topic: Mendel’s First Law of Heredity
Pages: 7–10
b. Give your reason why.
Ans: Waltzer mice crossed with normal mice produce all normal offspring, and it looks like the ratio of
phenotypes in crosses between these offspring (i.e., F2 mice) give normal to Waltzer in the ratio 3:1, as
expected from Mendel’s First Law.
Level: E
Topic: Mendel’s First Law of Heredity
Pages: 10
c. Perform a chi-square test to determine if the results given above are consistent with your hypothesis.
Ans:
Hypothesis: If Waltzer is caused by a recessive allele, expect 3:1 normal to Waltzer in F2 generation.
Chi-square formula 2 =  (obs – exp)2/exp
Test Obs Exp O-E (O-E)2 (O-E)2/E
124 129 5 25 0.194
48 43 5 25 0.581
 = 0.775 = 21df ~32% probability
Conclusion: Retain hypothesis; the probability of getting such observed numbers is high.

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Level: D
Topic: Mendel’s First Law of Heredity
Pages: 8–9
3. The figure below shows a small portion of a Huntington disease pedigree:
a. Using H as the symbol for the Huntington allele and h as the normal allele, give the genotypes at the
Huntington locus for each of the four children (A,B,C,D)
Ans: A = Hh B = hh C = hh D = Hh
Level: M
Topic: Huntington disease
Pages: 10–11
b. If child B marries an affected man, what is the probability that their first child would develop Huntington
disease?
Ans: 0.5 (or 50%)
Level: M
Topic: Huntington disease
Pages: 10–11
c. If their first child is unaffected, what is the probability that their second child is affected?
Ans: 0.5 (or 50%)
Level: M
Topic: Huntington disease
Pages: 10–11
d. If child A marries an unaffected woman and has two children, what is the probability that both children are
unaffected?
Ans: 0.5×0.5 = 0.25
Level: D
Topic: Huntington disease
Pages: 10–11
4. The figure below shows part of a PKU pedigree:
a. Using K as the symbol for the normal allele and k as the symbol for the PKU allele, give the possible
genotypes of the unaffected children (A,B,C).
Ans:
A = KK or Kk B = KK or Kk C= KK or Kk
Level: M
Topic: Phenylketonuria (PKU)
Pages: 11–12
b. Given that child A is an unaffected brother of a PKU child, what is the probability that he is a PKU carrier?
Ans: 0.66 (or 2/3)
Level: M

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Topic: Phenylketonuria (PKU)
Pages: 11–12
c. Assume A is a carrier. What genotype would his mate have to be for a child of the couple to be at risk for
PKU?
Ans: Kk (or kk, but this is less likely)
Level: E
Topic: Phenylketonuria (PKU)
Pages: 11–12
d. What are A’s chances (assuming he is a carrier) of marrying and producing children who are not at risk for
having PKU? (see Box 2.2, page 13 of text)
Ans: A cannot marry Kk or kk; mate has to be KK
Frequency of Kk = 2% Frequency of kk = 1/10000
So, frequency of KK = 1 – (0.2+ 0.00001) = 0.97999 around 98%
Level: D
Topic: Phenylketonuria (PKU)
Pages: 12–13
5. A mouse geneticist discovers two recessive alleles, each of which causes the mice to be deaf when two copies
are present.
a. How could the geneticist determine whether the alleles occur at the same locus (i.e., are alleles of the same
gene)?
Ans: Cross mice with two copies of one allele with mice having two copies of the other allele. If offspring are
deaf, then the alleles are at the same locus. If none of the mice are deaf, then the alleles are at separate loci.
Level: D
Topic: Mendel’s Second Law of Heredity
Pages: 14–17
b. If it is discovered that the alleles are at different loci, how could the geneticist discover if they were located
together on the same chromosome?
Ans: Do a test for linkage: Cross second-generation offspring (F2) from the mating of mice with both types of
deafness and look for a departure from the expected ratio of 9:3:3:1. A lack of recombinants would indicate
linkage.
Level: D
Topic: Mendel’s Second Law of Heredity
Pages: 14–17

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CHAPTER 3
Beyond Mendel’s Laws
VOCABULARY
Match the phrases to the terms below. Each term may be used once, more than once, or not at all.
Phrases
1.) Chromosomes that differ for males and females
2.) Describes genes on a sex chromosome
3.) Recessive alleles of this type produce disorders more common in males
4.) Describes a female that has an X-linked recessive allele for a disorder
5.) Having three copies of a chromosome instead of the usual two
6.) Process that produces incorrect numbers of chromosomes in cells
7.) Type of mutation caused by repeated three-nucleotide units
8.) Describes the situation where a disorder gets more severe as it is passed down generations
9.) Fragile X is caused by this type of mutation
10.) Describes a trait influenced by many genes
Terms
autosomes
carrier
correlation
first-degree relative
genetic anticipation
genomic imprinting
liability-threshold model
methylation
monosomy
mRNA
nondisjunction
nucleotide
polygenic
permutation
proband
qualitative trait
second-degree relative
sex chromosomes
transcription
translation
triplet repeat
trisomy
X-linked
Ans:

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1.) sex chromosomes 2.) X-linked 3.) X-linked 4.) carrier 5.) trisomy 6.) nondisjunction 7.) triplet
repeat 8.) genetic anticipation 9.) triplet repeat 10.) polygenic
Level: E
Topic: Beyond Mendel’s Laws
Pages: 20–38
PROBLEMS
1. Red-green color blindness is an X-linked recessive trait in humans. In four different families, the parents have
children together. For each of the families (A,B,C,D) shown in the table below, give a possible combination of
genotypes for the parents that could have resulted in the combinations of children found in each family. (Use
XC to indicate the color-blind allele and X to indicate the normal allele.)
Possible genotype:
Family Children Phenotype MOM DAD
A 2 males normal
2 females normal
B 2 females color-blind
3 females normal
2 males color-blind
C 4 males color-blind
4 females normal
D 2 females color-blind
2 males normal
Ans: There may be several possible answers for the genotypes of the parents, but given that the normal allele
is more common than the color-blindness allele, the following are the most likely genotypes:
Family A XX and XY; Family B XCX and XC;Y; Family C XCX and XY;
Family D XCX and XCY
Level: M
Topic: X-linked inheritance
Pages: 20–23
2. Draw a pedigree for Family B from the question above. Label the genotypes under the symbols for each
person.
Ans:
Level: M
Topic: X-linked inheritance
Pages: 20–23

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3. Down syndrome is caused by the presence of an extra copy of chromosome 21 and is also known as Trisomy
21.
a. Name the process that occurs in gamete formation to produce an abnormal number of chromosomes.
b. Name a prenatal test that a woman might need in order to detect the presence of genetic abnormalities.
c. Why is Down syndrome generally non-familial?
d. Why is it that mothers who are older than 40 have an increased risk of having a child with Down syndrome?
Ans:
a. non-disjunction
b. amniocentesis, chorionic villi sampling
c. Non-disjunction is a spontaneous chromosome mutation, and people with Down syndrome generally do not
reproduce, hence all occurrences are newly arisen.
d. The completion of cell division in females only takes place on ovulation (not continuously as in males) after
puberty, and the activation of immature eggs after many years gives rise to imperfect separation of
chromosomes.
Level: M
Topic: Chromosome mutations
Pages: 23–25
4. Huntington disease and fragile X syndrome are caused by a section of DNA that is repeated many times.
a. Explain how this type of mutation can give rise to
i. genetic anticipation
ii. premutation
b. What are the consequences of genetic anticipation and premutation in the families of those with Huntington
disease and fragile X?
Ans:
a. i. The expanded repeat section of DNA is unstable and may increase between each generation, resulting in
more severe cases in later generations. This phenomenon is known as genetic anticipation.
ii. A lower number of repeats than necessary to produce the disorder is recognized in relatives of those with
the disorder. This intermediate number of repeats between normal and full expansion is known as a
premutation.
b. For Huntington disease, genetic anticipation results in the appearance of HD at an earlier age and greater
severity as it is passed on. For fragile X, the symptoms become more severe as it is passed on. Premutation
carriers for both disorders are at greater than average risk of producing a child with the disorder.
Level: D
Topic: Expanded triplet repeats
Pages: 25–26
5. a. Look at the figures taken from the text, and explain how they show there is evidence that both schizophrenia
and general cognitive ability are influenced by genes.

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Figure 3.6 Risk for schizophrenia and genetic relatedness
Figure 3.7 Resemblance for general cognitive ability and genetic relatedness
b. Both correlations and morbidity risk estimates are measures that show familiality. Correlations are
generally more accurate measures of similarity between relatives. Why is it that correlations are given between
relatives for general cognitive ability but not for schizophrenia?
c. Continuous variation for a trait can be produced by many genes, each obeying Mendelian laws, and many of
the alleles are recognized to act in an “additive” manner. Explain what “additive” means in this context.
Ans:
a. Figure 3.6 shows the risk for schizophrenia increases the more genetically related someone is to a person
who has the disorder, and all relatives have an increased risk compared to population prevalence. Figure 3.7

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