Question 1

Briefly explain the different purposes for genetic screening in newborns and adults.

Accepted Answer

Genetic screening in newborns

Question 2

The human genome contains about ____ genes, which comprises about _____ of the genome.

Accepted Answer

A)  25,000; 2% 
B)  45,000; 2% 
C)  25,000; 50% 
D)  80,000; 2% 
E)  100,000; 1% 
A)  25,000; 2% 
B)  45,000; 2% 
C)  25,000; 50% 
D)  80,000; 2% 
E)  100,000; 1%

Question 3

Huntington's disease is caused by a:

Accepted Answer

A)  rare recessive allele 
B)  nucleotide triplet repeat 
C)  monosomic condition 
D)  dominant X-linked allele 
E)  deletion of part of chromosome 4 
A)  rare recessive allele 
B)  nucleotide triplet repeat 
C)  monosomic condition 
D)  dominant X-linked allele 
E)  deletion of part of chromosome 4

Question 4

Gene therapy involves:

Accepted Answer

A)  replacing all mutant alleles in body cells 
B)  replacing poor copies of alleles with normal alleles 
C)  replacing a mutant allele in certain body cells with a normal allele 
D)  repairing mutant alleles in certain body cells 
E)  improving gene expression in key body cells 
A)  replacing all mutant alleles in body cells 
B)  replacing poor copies of alleles with normal alleles 
C)  replacing a mutant allele in certain body cells with a normal allele 
D)  repairing mutant alleles in certain body cells 
E)  improving gene expression in key body cells

Question 5

Nondisjunction during mitosis leads to:

Accepted Answer

A)  an individual with a clone of abnormal cells 
B)  an individual with all abnormal cells 
C)  death of the daughter cells 
D)  nonviable sperm 
E)  normal daughter cells 
A)  an individual with a clone of abnormal cells 
B)  an individual with all abnormal cells 
C)  death of the daughter cells 
D)  nonviable sperm 
E)  normal daughter cells

Question 6

A genetic counselor will know that the probability of having a child affected with a condition inherited as an autosomal recessive would be:

Accepted Answer

A)  0.5 or 50% if both parents were homozygous normal 
B)  0.5 or 50% if both parents were heterozygous 
C)  0.25 or 25% if both parents were homozygous normal 
D)  0.25 or 25% if both parents were heterozygous 
E)  0% if both parents were heterozygous 
A)  0.5 or 50% if both parents were homozygous normal 
B)  0.5 or 50% if both parents were heterozygous 
C)  0.25 or 25% if both parents were homozygous normal 
D)  0.25 or 25% if both parents were heterozygous 
E)  0% if both parents were heterozygous

Question 7

According to data from the ENCODE project, approximately 80% of the human genome is associated with biological function.

Accepted Answer

A) True 
 B)False

Question 8

Cri du chat syndrome arises from:

Accepted Answer

A)  trisomy for chromosome 5 
B)  deletion of part of chromosome 5 
C)  a 14/21 reciprocal translocation 
D)  nondisjunction 
E)  a duplication of part of chromosome 15 
A)  trisomy for chromosome 5 
B)  deletion of part of chromosome 5 
C)  a 14/21 reciprocal translocation 
D)  nondisjunction 
E)  a duplication of part of chromosome 15

Question 9

Aneuploidies describe:

Accepted Answer

A)  a phenomenon that only occurs in plants 
B)  a condition in which an extra chromosome is present or one is absent 
C)  a defect that is always fatal in humans 
D)  an uncommon condition in humans 
E)  mutations that almost always have a beneficial effect on an individual 
A)  a phenomenon that only occurs in plants 
B)  a condition in which an extra chromosome is present or one is absent 
C)  a defect that is always fatal in humans 
D)  an uncommon condition in humans 
E)  mutations that almost always have a beneficial effect on an individual

Question 10

In a chromosomal inversion, a segment of a chromosome is:

Accepted Answer

A)  reversed 
B)  duplicated 
C)  lost 
D)  attached to a nonhomologous chromosome 
E)  deleted 
A)  reversed 
B)  duplicated 
C)  lost 
D)  attached to a nonhomologous chromosome 
E)  deleted

Question 11

An individual who inherits the specific chromosome ____ deletion from his/her ____ will most likely have Prader-Willi syndrome rather than Angelman syndrome.

Accepted Answer

A)  15; mother 
B)  15; father 
C)  21; mother 
D)  21; father 
E)  12; father 
A)  15; mother 
B)  15; father 
C)  21; mother 
D)  21; father 
E)  12; father

Question 12

Figure 16-2
 
The inheritance pattern demonstrated by the pedigree in accompanying figure is:

Accepted Answer

A)  autosomal recessive 
B)  autosomal dominant 
C)  X-linked recessive 
D)  X-linked dominant 
E)  undeterminable from these data 
A)  autosomal recessive 
B)  autosomal dominant 
C)  X-linked recessive 
D)  X-linked dominant 
E)  undeterminable from these data

Question 13

Turner syndrome is an example of a __________ condition.

Accepted Answer

A)  monosomic 
B)  disomic 
C)  trisomic 
D)  polyploid 
E)  fragile site 
A)  monosomic 
B)  disomic 
C)  trisomic 
D)  polyploid 
E)  fragile site

Question 14

A karyotype reveals that an individual is XYY. Based on your knowledge of human genetics, you correctly conclude that this individual has ____, and is a phenotypically _____.

Accepted Answer

A)  Turner syndrome; sterile female 
B)  Turner syndrome; fertile female 
C)  Klinefelter syndrome; sterile male 
D)  Klinefelter syndrome; fertile male 
E)  XXY karyotype; fertile male 
A)  Turner syndrome; sterile female 
B)  Turner syndrome; fertile female 
C)  Klinefelter syndrome; sterile male 
D)  Klinefelter syndrome; fertile male 
E)  XXY karyotype; fertile male

Question 15

The science of comparative genomics to study human gene function is possible because much of the human genome is conserved among other closely related species.

Accepted Answer

A) True 
 B)False

Question 16

Comparing the human and mouse genomes has revealed that the mouse and human:

Accepted Answer

A)  share few genes 
B)  genomes are identical 
C)  genomes have no genes in common 
D)  share most genes 
E)  genomes have identical noncoding DNA 
A)  share few genes 
B)  genomes are identical 
C)  genomes have no genes in common 
D)  share most genes 
E)  genomes have identical noncoding DNA

Question 17

In individuals with cystic fibrosis:

Accepted Answer

A)  the mucus enhances the action of the cilia that line the bronchi. 
B)  the mucus prevents the growth of dangerous bacteria in the respiratory system. 
C)  the mucus in the respiratory system is thinner than in unaffected individuals. 
D)  abnormal mucous secretions occur throughout the body systems. 
E)  abnormal mucous secretions inhibit infections by bacteria. 
A)  the mucus enhances the action of the cilia that line the bronchi. 
B)  the mucus prevents the growth of dangerous bacteria in the respiratory system. 
C)  the mucus in the respiratory system is thinner than in unaffected individuals. 
D)  abnormal mucous secretions occur throughout the body systems. 
E)  abnormal mucous secretions inhibit infections by bacteria.

Question 18

The normal human karyotype contains ____ total chromosomes, consisting of ____ autosomes and ____ sex chromosomes.

Accepted Answer

A)  23; 22; 1 
B)  44; 22; 1 
C)  46; 22; 2 
D)  46; 23; 0 
E)  22; 20; 2 
A)  23; 22; 1 
B)  44; 22; 1 
C)  46; 22; 2 
D)  46; 23; 0 
E)  22; 20; 2

Question 19

Autosomal aneuploidies arise by:

Accepted Answer

A)  chromosome breakage and rejoining 
B)  meiotic nondisjunction 
C)  errors in crossing-over 
D)  mistakes in chromosome replication 
E)  mitotic nondisjunction 
A)  chromosome breakage and rejoining 
B)  meiotic nondisjunction 
C)  errors in crossing-over 
D)  mistakes in chromosome replication 
E)  mitotic nondisjunction

Question 20

What explains the viability of individuals with aneuploidies for the sex chromosomes?

Accepted Answer

A)  Sex chromosomes carry less information than autosomes. 
B)  The lack of crossing-over between the X and the Y chromosomes. 
C)  Sex chromosomes are smaller than most autosomes. 
D)  The information on sex chromosomes is duplicated on the autosomes. 
E)  Dosage compensation, by which one X chromosome per cell is inactivated. 
A)  Sex chromosomes carry less information than autosomes. 
B)  The lack of crossing-over between the X and the Y chromosomes. 
C)  Sex chromosomes are smaller than most autosomes. 
D)  The information on sex chromosomes is duplicated on the autosomes. 
E)  Dosage compensation, by which one X chromosome per cell is inactivated.

Exam 16: Human Genetics and the Human Genome

Briefly explain the different purposes for genetic screening in newborns and adults.

The human genome contains about ____ genes, which comprises about _____ of the genome.

Huntington's disease is caused by a:

Gene therapy involves:

Nondisjunction during mitosis leads to:

A genetic counselor will know that the probability of having a child affected with a condition inherited as an autosomal recessive would be:

According to data from the ENCODE project, approximately 80% of the human genome is associated with biological function.

Cri du chat syndrome arises from:

Aneuploidies describe:

In a chromosomal inversion, a segment of a chromosome is:

An individual who inherits the specific chromosome ____ deletion from his/her ____ will most likely have Prader-Willi syndrome rather than Angelman syndrome.

Figure 16-2 The inheritance pattern demonstrated by the pedigree in accompanying figure is:

Turner syndrome is an example of a __________ condition.

A karyotype reveals that an individual is XYY. Based on your knowledge of human genetics, you correctly conclude that this individual has ____, and is a phenotypically _____.

The science of comparative genomics to study human gene function is possible because much of the human genome is conserved among other closely related species.

Comparing the human and mouse genomes has revealed that the mouse and human:

In individuals with cystic fibrosis:

The normal human karyotype contains ____ total chromosomes, consisting of ____ autosomes and ____ sex chromosomes.

Autosomal aneuploidies arise by:

What explains the viability of individuals with aneuploidies for the sex chromosomes?

A View of Life

Atoms and Molecules: the Chemical Basis of Life

The Chemistry of Life: Organic Compounds

Organization of the Cell

Biological Membranes

Cell Communication

Energy and Metabolism

How Cells Make Atp: Energy-Releasing Pathways

Photosynthesis: Capturing Light Energy

Chromosomes, Mitosis, and Meiosis

The Basic Principles of Heredity

DNA: The Carrier of Genetic Information

Gene Expression

Gene Regulation

DNA Technology and Genomics

Developmental Genetics

Introduction to Darwinian Evolution

Evolutionary Change in Populations

Speciation and Macroevolution

The Origin and Evolutionary History of Life

The Evolution of Primates

Understanding Diversity: Systematics

Viruses and Subviral Agents

Bacteria and Archaea

Protists

Seedless Plants

Seed Plants

The Fungi

An Introduction to Animal Diversity

Sponges, Cnidarians, Ctenophores, and Protostomes

The Deuterostomes

Plant Structure, Growth, and Development

Leaf Structure and Function

Stem Structure and Transport

Roots and Mineral Nutrition

Reproduction in Flowering Plants

Plant Developmental Responses to External and Internal Signals

Animal Structure and Function: an Introduction

Protection, Support, and Movement

Neural Signaling

Neural Regulation

Sensory Systems

Internal Transport

The Immune System: Internal Defense

Gas Exchange

Processing Food and Nutrition

Osmoregulation and Disposal of Metabolic Wastes

Endocrine Regulation

Reproduction

Animal Development

Animal Behavior

Introduction to Ecology: Population Ecology

Community Ecology

Ecosystems and the Biosphere

Ecology and the Geography of Life

Biological Diversity and Conservation Biology

Filters

The human genome contains about genes, which comprises about _ of the genome.

An individual who inherits the specific chromosome deletion from his/her will most likely have Prader-Willi syndrome rather than Angelman syndrome.

A karyotype reveals that an individual is XYY. Based on your knowledge of human genetics, you correctly conclude that this individual has , and is a phenotypically _.

The normal human karyotype contains total chromosomes, consisting of autosomes and sex chromosomes.