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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

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    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

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    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

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    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

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    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 =  2 1df ~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 X C 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 X C X and X C; Y; Family C X C X and XY; Family D X C X and X C Y 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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