Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 1: DNA Structure and Function Multiple Choice Identify the choice that best completes the statement or answers the question.

1. In which body or cell area are most genes in humans located?

A. Nucleus B. Mitochondrion C. Cytoplasm D. Plasma membrane 2. Which condition or statement exemplifies the concept of genomics rather than genetics?

A. The gene for insulin is located on chromosome 11 in all people.

B. Expression of any single gene is dependent on inheriting two alleles.

C. Sex-linked recessive disorders affect males more often than females.

D. One allele for each gene is inherited from the mother, and one is inherited from

the father.

3. What is the purpose of phosphorous in a DNA strand?

A. Linking the nucleotides into a strand B. Holding complementary strands together C. Ensuring that a purine is always paired with a pyrimidine D. Preventing the separation of double-stranded DNA into single-stranded DNA 4. What is the term used to define alternative forms of a gene that may result in different expression

of the trait coded for by that gene?

A. Alleles B. Bases C. Centromeres D. Diploids 5. What percentage of bases in a stretch of double-stranded DNA that contains 30% guanine (G)

bases would be adenine (A)?

A. 70% B. 60% C. 30% D. 20% 6. What is the term used to describe the organized picture of the paired chromosomes within a cell

used to determine whether chromosome numbers, structures, and banding patterns are normal?

A. Pedigree B. Phenotype C. Karyotype D. Autotype 7. What would be the sequence of DNA that is complementary to a DNA section with the

base sequence of GGTCAATCCTTAG? A. GATTCCTAACTGG

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. TTGACCGAAGGCT C. AACTGGCTTCCGA D. CCAGTTAGGAATC 8. Which of these complementary base pairs form the strongest or “tightest” association?

A. Adenine and thymine B. Cytosine and guanine C. Guanine and thymine D. Cytosine and adenine 9. What activity occurs during M phase of the cell cycle?

A. The cell undergoes cytokinesis.

B. Activity stops, and the cell “sleeps.”

C. All DNA is completely replicated.

D. The cell greatly increases protein synthesis.

10. Which chromosome number represents the euploid state for normal human somatic

cells?

A. 44 B. 46 C. 47 D. 48 11. How does the proteome differ from the genome?

A. The proteome changes in response to intracellular and extracellular signals.

B. The genome changes in response to intracellular and extracellular signals.

C. The proteome is stable in somatic cells and unstable in germ cells, whereas the

genome is stable in both somatic cells and germ cells.

D. The genome is stable in somatic cells and unstable in germ cells, whereas the

proteome is stable in both somatic cells and germ cells.

12. What is the most outstanding feature of a mature haploid cell?

A. It is usually homozygous.

B. The sex chromosomes are missing.

C. Only one chromosome of each pair is present.

D. DNA synthesis occurs after mitosis instead of before.

13. At what phase of the cell cycle are chromosomes visible as separate structures?

A. G1 B. G2 C. S D. M 14. Which statement about the cell cycle phase of G0 is true?

A. Hyperplastic growth in place of hypertrophic growth B. Performance of specific differentiated functions C. Initiation and completion of nucleokinesis D. Replication of DNA 15. What is the result of normal DNA replication?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Formation of two new daughter cells B. Formation of two identical sets of DNA C. Disappearance of the original parent cell D. Activation and attachment of spindle fibers 16. Which statement regarding chromosome structure or function is true?

A. The chromatids of any single chromosome are known as “sister chromatids.”

B. The genes located on the telomeres of chromosomes are identical to the genes

in the centromeres.

C. Immediately before the mitosis phase of cell division, the chromosomes of

all somatic cells are haploid.

D. A specific gene allele on one chromosome has a complementary allele on the

other chromosome of a pair.

17. Why does a person with normal chromosomes only have two alleles for any single gene trait?

A. A minimum of two alleles is required for the expression of monogenic traits.

B. When a dominant allele is paired with a recessive allele, only the dominant allele

is expressed, and the recessive allele is silent.

C. One allele for the monogenic trait is on the paternally derived chromosome,

and the other allele is on the maternally derived chromosome.

D. Expression of more than two alleles of any single-gene trait results in enhanced expression of recessive alleles and suppressed expression of dominant alleles.

18. Under what normal condition are genotype and phenotype always the same?

A. Euploidy of alleles B. Aneuploidy of alleles C. Homozygosity of alleles D. Heterozygosity of alleles 19. What would be the expected result of a drug that affected a particular tissue by causing new DNA

to form with covalent bonds instead of hydrogen bonds?

A. None of the cells in the affected tissue would be able to leave G0 and enter the cell cycle.

B. Replication of DNA would result in identical DNA strands instead of complementary strands.

C. Mitosis of cells in the tissue would result in the production of three new daughter cells instead of just two.

D. The new cells that formed within this tissue would not be able to complete the next round of mitosis successfully.

20. How does the DNA enzyme topoisomerase contribute to DNA replication?

A. Unwinds the double helix and separates the double-stranded DNA B. Creates a “nick” in the DNA supercoils, allowing them to straighten

before replication C. Initiates DNA synthesis in multiple sites down the strand, making the process

more efficient D. Connects and links the individual pieces of newly synthesized DNA to form

a single strand

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

21. Where is telomeric DNA located?

A. At the tips of the p and q arms of chromosomes.

B. In the mitochondria of all somatic cells C. Only in the germ cells (ova and sperm) D. Within the histones of the solenoid 22. What is the purpose of a chromosome centromere?

A. Connecting sister chromatids to form a chromosome B. Preventing the chromosome arm tips from unraveling C. Allowing chromatids to separate during DNA replication D. Ensuring that DNA replication proceeds only in the 3'-to-5' direction 23. Which genetic process would be disrupted in one cell if it could not form chromosomes?

A. DNA replication B. Gene-directed protein synthesis C. Delivery of genetic information to new cells D. Conversion of a nucleoside into a nucleotide 24. What are the expected expressed blood types of children born to a mother who is B/O for blood

type and a father who is A/B for blood type?

A. 25% A, 25% B, 25% O, 25% AB B. 25% A, 50% B, 0% O, 25% AB C. 50% A, 25% B, 25% O, 0% AB D. 50% A, 25% B, 0% O, 25% AB 25. A person’s karyotype shows 44 autosomes and one X chromosome. What is the best interpretation

of this karyotype?

A. The karyotype is aneuploid, and the individual has only one allele for each of the genes on the X chromosome.

B. The karyotype is aneuploid, and the individual is experiencing the pathologic condition of haploidy.

C. The karyotype is euploid, making the individual a genotypic female and a phenotypic male.

D. The karyotype is euploid, making the individual a genotypic male and a phenotypic female.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 1: DNA Structure and Function Answer Section MULTIPLE CHOICE 1. ANS: A

Most genes are part of the DNA located in the nucleus of body cells. Only a few genes are located in a cell’s mitochondrion or mitochondria. There are no genes or DNA in either the cytoplasm or the plasma membranes of any cell.

PTS: 1 2. ANS: A

Genetics is the study of the general mechanisms of heredity and the variation of inherited traits.

Genomics is the study of the function of all the nucleotide sequences present within the entire genome of a species, including genes in DNA coding regions and DNA noncoding regions.

Selections B, C, and D all refer to mechanisms of heredity. Only selection A refers to the function of a specific nucleotide sequence.

PTS: 1 3. ANS: A

Each nucleoside becomes a complete nucleotide when a phosphate group is attached. The phosphates have multiple binding sites, and each one can link to two nucleotides. These linkages allow the nucleotides to be connected when placed into the DNA strand. The nucleotides within each strand are held in position by the linked phosphate groups, which act like the string holding a strand of beads together, forming a necklace.

PTS: 1 4. ANS: A

For each single gene, two alternative forms of that gene, known as alleles, together control how that gene is expressed. The alleles may be identical in their sequence but do not have to be. When a dominant allele is paired with a recessive allele, only the dominant allele is expressed, and the recessive allele is silent. When a dominant allele is paired with another dominant allele, they are both expressed (usually equally). Recessive alleles are only expressed when they are homozygous.

Bases are the essential part of a nucleotide, of which there are many within any gene region.

Centromeres are the pinched-in part of a chromosome between the p arms and the q arms. The term diploid refers to the normal number of chromosome pairs within a cell. It is an adjective, not a noun.

Therefore, the plural diploids does not exist.

PTS: 1 5. ANS: D

Because of complementary pairing, if 30% of the bases are guanine (G), which always pairs with cytosine C, these two bases account for 60% of the total bases in this stretch. The remaining bases make up 40% of the total. This 40% is composed of equal percentages of thymine (T) and adenine (A).

PTS: 1 6. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A karyotype is a picture of an organized arrangement of all of the chromosomes within one cell during the metaphase section of mitosis. The chromosomes are paired and then arranged by number according to size and centromere position. The banding pattern of each pair is analyzed to determine whether areas have been deleted, expanded, or translocated. A pedigree also is a picture, but it illustrates several generations of a family history. Phenotypes are observable traits.

Autotype is not a genetic term.

PTS: 1 7. ANS: D

Because doubled stranded DNA (ds-DNA) is complementary in that A always pairs with T and G always pairs with C, wherever a G is located on strand 1, the complementary base in the same position on strand 2 is C; wherever a C is located on strand 1, the complementary base in the same position on strand 2 is G; wherever a T is located on strand 1, the complementary base in the same position on strand 2 is A; and wherever an A is located on strand 1, the complementary base in the same position on strand 2 is T. PTS: 1 8. ANS: B

Cytosine and guanine normally pair together, whereas adenine and thymine normally pair together. The reason for this specific and complementary pairing of bases is related to the forces that hold the two DNA strands together. The two strands are held loosely together, most of the time, by weak hydrogen bonds. Within a base pair, the hydrogen bonds form between the two nucleotides. Adenine and thymine each have a site for two hydrogen bonds to form, whereas cytosine and guanine each have three sites for hydrogen bonds to form. With three hydrogen bonds instead of just two, the cytosine–guanine pair is tighter (stronger) than an adenine–thymine pair. Note, a purine must always pair with a pyrimidine: they each can only pair with the base that can form the same number of hydrogen bonds. Adenine and cytosine do not pair, and neither do guanine and thymine.

PTS: 1 9. ANS: A

The M stands for mitosis. This is the time during the cell cycle after the DNA has successfully replicated completely into two identical sets of double-stranded DNA during S phase, and then progressed through G2 phase into M phase. During M phase, the chromatids separate, each half goes into a new separate nucleus (nucleokinesis), and the cell has two nuclei. Just after nucleokinesis, the cell pulls apart into two cells (cytokinesis), each with a complete set of chromosomes in the nucleus. These two new daughter cells are identical to the parent cell that initiated cell division. There is no “sleep” stage of the cell cycle, although there is a reproductive resting state (G0) outside of the cell cycle.

PTS: 1 10. ANS: B

Ploidy is the actual number of chromosomes present in a single cell nucleus at mitosis. Humans have 46 chromosomes divided into 23 pairs. When the nucleus contains both pairs of all chromosomes, the number present is the diploid chromosome number (2N). Normal human somatic cells (body cells that are not reproductive cells) with a nucleus have the diploid number of chromosomes, 23 pairs. When a cell’s nucleus contains the normal diploid number of chromosomes for the species, the cell is termed euploid.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 11. ANS: A

The proteome is the protein content of any given cell. It is dynamic, changing in response to intracellular and extracellular (environmental) signals. Therefore, the proteome is not an exact copy of the genome. The genome is stable in all cells, and the proteome expresses different proteins (both enzymes and structural proteins) at different times in different cells.

PTS: 1 12. ANS: C

A haploid cell has a nucleus that contains only half of each chromosome pair, 23 chromosomes (1N). In a normal haploid cell, one sex chromosome is present. Because there is only one chromosome of each pair present, only one allele of each gene is present, and the cell is not homozygous. A mature haploid cell does not undergo cell division (mitosis) and does not replicate its DNA. PTS: 1 13. ANS: D

A chromosome is a temporary but consistent state of condensed DNA structure formed for the purpose of cell division during metaphase of mitosis (M phase). At other times in the cell cycle, the DNA is so loosely coiled as a double helix that the basic structure of a chromosome is not visible with a standard microscope. The DNA can be seen with an electron microscope as a double helix, not a chromosome, during G1, G2, and S phases.

PTS: 1 14. ANS: B

G0 is a reproductive resting state of a cell outside of cell division in which the cell performs its normal differentiated functions. Therefore, nucleokinesis (a process of M phase of the cell cycle) and replication of DNA (a process of S phase of the cell cycle) do not occur. Hyperplastic growth is growth by cell division, which does not occur during the reproductive resting state of G0.

PTS: 1 15. ANS: B

In S phase of the cell cycle, the DNA replicates completely into two identical sets of doublestranded DNA. This occurs in preparation for cell division, which then generates two new daughter cells.

Spindle fiber formation, activation, and attachment are critical for mitosis but have no role in DNA replication. The process of DNA replication occurs inside the original parent cell that is undergoing mitosis but does not make the parent cell disappear.

PTS: 1 16. ANS: A

Each longitudinal left and right half of any one chromosome is a chromatid. The two chromatids of a chromosome are homologous and termed sister chromatids. Gene alleles on separate chromosomes of a pair are not considered “complementary”; only gene sequences are complementary.

PTS: 1 17. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

For each single gene at a specific chromosome location, two alleles together control how that gene is expressed even for single-gene traits that have more than two possible alleles. Regardless of how many different possible alleles are present in the entire human population, each person only has two because he or she has only two chromosomes (one inherited from his or her father and one inherited from his or her mother) per pair with one allele on each chromosome.

Selection B is true but does not answer the question asked. Only two alleles for each monogenic trait should be present. Both are not required for the expression of all monogenic traits.

PTS: 1 18. ANS: C

Homozygous alleles are identical on both chromosomes of a pair and result in the expression of both of these identical alleles in the phenotype, regardless of whether the alleles are both recessive or both dominant. Heterozygous alleles are different on both chromosomes of a pair. Although they may both be expressed or one may be silent, the phenotype does not always reflect the genotype. Trisomy of alleles would require an extra chromosome. This condition, unless the alleles are homozygous, does not always result in identical genotype and phenotype. Triploidy, like trisomy, would require an extra chromosome copy (actually, a whole extra set of chromosomes).

This condition, unless the alleles are homozygous, does not always result in identical genotype and phenotype.

PTS: 1 19. ANS: D

If, during S phase of one cell cycle, the replicated DNA had covalent bonds rather than hydrogen bonds, the cell could complete that round of cell division, resulting in two new daughter cells.

When either of these two new daughter cells then reentered the cell cycle, it could not complete the cycle. The tight covalent bonds of the DNA in each of these cells would not break apart to allow each strand to separate and be used as templates for DNA replication and synthesis. Thus, the cycle would be arrested at this stage.

PTS: 1 20. ANS: B

Replication of DNA is performed when double-stranded DNA is separated and each strand is used as a template to guide the correct construction of a complementary strand. Synthesis of new DNA begins at multiple sites, creating many short pieces. The enzyme DNA ligase connects or links the individual pieces of newly synthesized DNA during replication, forming a new single strand complementary to the template strand. The enzyme DNA helicase unwinds the double helix and initially separates the ds-DNA. The group of topoisomerase enzymes creates a “nick” in the supercoils of ds-DNA, allowing them to loosen so that eventually the two strands can separate.

The enzyme primase is responsible for initiating DNA synthesis in multiple sites down the single strand being copied.

PTS: 1 21. ANS: A

Telomeric DNA is a type of noncoding DNA located at the tips of chromosomes. Its purpose is to prevent DNA strands from unraveling. It is present in all linear DNA. The mitochondria do not contain linear DNA or true chromosomes. It is present in all somatic cells, as well as germ cells.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

22. ANS: A

A centromere is the pinched-in area of the chromosome that connects the two halves (sister chromatids) together in a metaphase chromosome. It also connects the chromosome segments above it and below it. The centromere has no role in DNA replication. The telomeres prevent chromosome unraveling. DNA replication proceeds in the 5'-to-3' direction and is determined by specific enzymes, not the centromere.

PTS: 1 23. ANS: C

Chromosomes are temporary structures that have the important job of making the delivery of DNA to the two new cells precise so that one new cell does not get more or less than the correct amount of DNA and the correct distribution of the genes. This precision is critical for the new cells to be able to function normally. Chromosomes are not involved in either DNA synthesis or protein synthesis.

PTS: 1 24. ANS: B

Blood type alleles A and B are codominant, and O is recessive. Thus when O is paired with either A or B, it is not expressed. When A and B are paired together, both are expressed, and the blood type is AB. In this situation both parents are heterozygous. For each pregnancy there is one chance out of four (25%) of the AB blood type being expressed, two chances out of four (50%) of the B blood type being expressed (with either a B/B genotype or a B/O genotype), one chance out of four (25%) of the A blood type being expressed (from an A/O genotype), and no chance (0%) of an O blood type (O/O genotype) being expressed.

PTS: 1 25. ANS: A

A euploid karyotype for humans has a total of 46 chromosomes, 44 autosomes, and a pair (2) of sex chromosomes. The fact that this karyotype has only 45 chromosomes makes it aneuploid.

Because the person is missing a sex chromosome and only one X is present, the person is female but has only one allele for each gene on the X chromosome (and none of the alleles for genes on the Y chromosome).

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 2: Protein Synthesis Multiple Choice Identify the choice that best completes the statement or answers the question.

1. What is the relationship among genes, DNA, and proteins?

A. DNA is composed of a series of amino acids that provide the directions

for synthesizing proteins.

B. Protein is composed of DNA that is organized into specific gene sequences

called amino acids.

C. A gene is a section of DNA that provides the directions for synthesizing a

specific protein.

D. Proteins are the nitrogenous bases that form double strands of DNA in its

helical shape.

2. What is the best meaning for the term gene expression?

A. The location of a specific gene allele on a specific autosomal chromosome B. The specific trait or protein coded for by a single gene is actually present C. The ability of a single gene to code for more than one trait or characteristic D. The loss of a trait or characteristic from one family generation to the

next generation 3. What is the difference between DNA transcription for DNA synthesis and DNA transcription

for protein synthesis?

A. Transcription for DNA synthesis is rapidly followed by the process of translation.

B. Transcription for protein synthesis has “greater fidelity” than does transcription for DNA synthesis.

C. Transcription for protein synthesis occurs only in cells undergoing mitosis, and transcription for DNA synthesis occurs in both dividing and nondividing cells.

D. Transcription for DNA synthesis occurs with both the “sense” and the “antisense” strands, while transcription for protein synthesis occurs with only the “antisense” strand.

4. Which mature messenger RNA strand correctly reflects the accurate transcription of the

following segment of DNA, in which large letters represent introns and small letters represent exons? tTGCGaAccaGaCTtaaAAtTAAA A. AUGGUUAUUA B. ACGCTCGATTATTT C. CGCUCGAUUAUUU D. AACGCUUGGUCUGAAUUUUAAUUU 5. What is the function of ribosomes (also known as ribosomal RNA) in protein synthesis?

A. Allow interpretation of the two strands of DNA to determine which is the

“sense” strand and which is the “antisense” strand B. Serve as the coordinator mechanism to allow proper reading of the mRNA

and placement of the correct amino acid in the sequence by the tRNAs C. Allow further processing of synthesized proteins (posttranslational modification)

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

in order to ensure that the final product is physiologically active D. Serve as transport molecules able to move a specific amino acid to the site of protein synthesis (peptide chain elongation) in the correct sequence 6. A strand of recently transcribed mRNA contains the following components: intron (1), intron (2),

exon (3), intron (4), exon (5), exon (6), exon (7), intron (8). Which sequence is expected to appear in the mature mRNA? A. 1, 2, 3, 4, 5, 6, 7, 8 B. 2, 3, 4, 5, 6, 7 C. 1, 2, 4, 8 D. 3, 5, 6, 7 7. Which process occurs outside of the nucleus?

A. DNA transcription B. RNA transcription C. Splicing out of introns D. Translation of mRNA 8. What would be the consequence for protein synthesis if only limited amounts of adenine were

available in a cell?

A. Increased rate of mRNA degradation B. Increased formation of mutation “hot spots” C. Decreased production of cellular proteins D. Decreased amounts of uracil in the cytoplasm 9. Which process would be directly inhibited by a lack of conversion of thymine to uracil?

A. Translation B. Transcription C. MicroRNA silencing D. Posttranscriptional modification 10. What would be the sequence of RNA complementary to single-stranded DNA with the base

sequence of ACCTGAACGTCGCTA? A. TGGACTTGCAGCGAT B. ACCTGAACGTCGCTA C. UGGACUUGCAGCGAU D. ACCUGAACGUCGCUA 11. Which events, structures, or processes are likely to trigger transcription of the beta-globin gene?

A. Anemia and TATA boxes upstream from the beta-globin gene B. Anemia and polyadenylation downstream from the beta-globin gene C. Polycythemia and TATA boxes upstream from the beta-globin gene D. Polycythemia and polyadenylation downstream from the beta-globin gene 12. After a protein is synthesized during translation, what further process or processes is/are needed

for it to be fully functional?

A. No further processing beyond the linear arrangement of amino acids is required.

B. Although minimal function can occur in the linear form, the protein is more active when it undergoes mitosis.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

C. The protein first twists into a secondary structure and then “folds” into a specific

tertiary structure for activation and function.

D. The initial protein produced is a “preprotein” that requires a series of

depolarizations by electrical impulses for conversion to an active protein.

13. How does an “anticodon” participate in protein synthesis?

A. Splicing out the introns to form a functional and mature messenger RNA B. Identifying which DNA strand is the “sense” strand to transcribe into RNA C. Ensuring the appropriate tRNA places the correct amino acid into the protein D. Interpreting the correct “stop” triplet or codon that signals for

translation termination 14. The protein glucagon contains 29 amino acids in its active linear form. What is the minimum

number of bases present in the mature messenger RNA for this protein?

A. 29 B. 58 C. 87 D. 116 15. Which feature or characteristic is most critical for protein function or activity?

A. The number of amino acids B. The sequence of amino acids C. Deletion of all active exons D. Transcription occurring after translation 16. How does a “codon” participate in protein synthesis?

A. Carrying amino acid for peptide bond attachment B. Ensuring that ribosomal RNA is securely wrapped around the mature mRNA C. Preventing microRNA from binding to mRNA and prematurely degrading it D. Indicating which amino acid is to be placed within the growing protein chain 17. How does replacement of thymine with uracil in messenger RNA help in the process of

protein synthesis?

A. Allowing messenger RNA to leave the nucleus B. Ensuring only the “antisense” strand of DNA is transcribed C. Determining the placement of the “start” signal for translation D. Promoting posttranslational modification for conversion to an active protein 18. How does the process of polyadenylation affect protein synthesis?

A. Binding to the antisense DNA strand to prevent inappropriate transcription B. Promoting attachment of ribosomes to the correct end of messenger RNA C. Linking the exons into the mature messenger RNA D. Signaling the termination of mRNA translation 19. Why are ribonucleases that digest mature messenger RNA a necessary part of protein synthesis?

A. These enzymes prevent overexpression of critical proteins.

B. Without ribonucleases, messenger RNA could leave one cell type and lead to

excessive protein synthesis in a different cell type.

C. When ribonucleases degrade RNA, the degradation products are recycled, making

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

protein synthesis more energy efficient.

D. The activity of these enzymes promotes increased translation of individual messenger RNAs so that fewer RNA molecules are needed for protein production.

20. Which statement about the introns within one gene is correct?

A. These small pieces of DNA form microRNAs that regulate gene expression.

B. They are part of the desert DNA composing the noncoding regions.

C. When expressed, they induce posttranslational modifications.

D. The introns of one gene may be the exons of another gene.

21. Which DNA segment deletion would cause a frameshift mutation?

A. TCT B. GAGTC C. TACTAC D. GCATGACCC 22. A person who is worried that he may have inherited the gene mutation for Huntington disease is

told that he has the “wild-type” form of this gene. What is the best interpretation of this finding?

A. His gene for Huntington disease (HD) has more “hot spots” for mutations than the general population.

B. His Huntington disease has unusual mutations of unknown significance.

C. His Huntington disease gene is considered normal.

D. He has no Huntington disease gene.

23. What is the expected result of a “nonsense” point mutation?

A. Total disruption of the gene reading frame, no production of protein B. Replacement of one amino acid with another in the final gene product C. Replacing an amino acid codon with a “stop” codon, resulting in a truncated

protein product D. No change in amino acid sequence and no change in the composition of the protein

product 24. What makes a frameshift mutational event more serious than a point mutational event?

A. Frameshift mutations occur primarily in germline cells, and point mutations occur

only in somatic cells.

B. Frameshift mutations result in the deletion or addition of whole chromosomes

(aneuploidy), and point mutations are undetectable at the chromosome level.

C. The rate of frameshift mutations increases with aging because DNA repair

mechanisms decline, whereas the rate of point mutations is unchanged with age.

D. When the mutations occur in expressed genes, frameshift mutations always result in disruption of the gene function, whereas a point mutation can be silent.

25. What is the expected outcome when a person (twin A) experiences a large deletion of DNA in one

of his noncoding regions and his monozygotic twin (twin B) does not?

A. DNA identification of each twin will be more specific.

B. Only their somatic cells will remain identical at all loci.

C. Only their germline cells will remain identical at all loci.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

D. They will now be dizygotic twins instead of monozygotic twins.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

26. Which statement about single-nucleotide polymorphisms (SNPs) is true?

A. SNPs can change an exon sequence into an intron sequence.

B. SNPs can change an intron sequence into an exon sequence.

C. SNPs are generally responsible for frameshift mutations.

D. SNPs are generally responsible for point mutations.

27. Why are people who have poor DNA repair mechanisms at greater risk for cancer development?

A. Their cancers are usually resistant to chemotherapy.

B. Their somatic mutations are more likely to be permanent.

C. They have greater exposure to environmental carcinogens.

D. They have sustained a mutational event in all cells and tissues.

28. How does an acquired mutation in a somatic cell gene leading to cancer development affect a

person’s ability to pass on a predisposition for that cancer type to his or her children?

A. The predisposition can only be passed on if the person with the somatic cell mutation is female.

B. There is no risk of passing on a cancer predisposition to one’s children from a somatic cell mutation.

C. The risk for predisposition is dependent on which tissue type experienced the somatic mutation.

D. Multiple somatic mutations are required for passing on a predisposition to cancer development.

29. Which factor has the greatest influence on protein tertiary structure?

A. The presence of a poly-A tail B. The specific amino acids that are in close proximity to each other C. Bond formation between amino acids that are distant from each other D. The number and position of additional proteins needed to form the

complex structure 30. Jack and Jill go up a hill that has high levels of gamma radiation emission. Jack suffers 10 point

mutational events in a noncoding region, and Jill suffers only one frameshift mutation in the insulin gene–coding region of all her pancreatic beta cells. What are the possible and probable outcomes of these events for both people?

A. Jack will have major deficiencies in the production of 10 proteins; Jill will have reduced insulin activity.

B. Jack will have less functional proteins and an increased risk for cancer; Jill will have type 2 diabetes mellitus.

C. Jack will have few, if any, effects on protein synthesis but will have more personal DNA markers; Jill will not produce any functional insulin and will have type 1 diabetes mellitus.

D. Jack will not have any change in protein synthesis or function; Jill will have an increased risk for developing type 1 diabetes mellitus and can pass this risk on to her children.

31. A new experimental drug has been developed that reduces the activity of microRNA in the beta

cells of the pancreas of people with type 2 diabetes mellitus. If this drug was specific only for pancreatic beta cells, what would be its effect?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Increased production of insulin B. Decreased production of insulin C. Increased degradation of insulin D. Decreased degradation of insulin

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 2: Protein Synthesis Answer Section MULTIPLE CHOICE 1. ANS: C

The correct sequence and relationships are listed in option C. A gene is a section of a specific DNA sequence that encodes the instructions for the amino acid sequence of a specific protein.

The DNA is “read” and transcribed into messenger RNA, which is translated as a series of amino acids. When these amino acids are joined together in the correct sequence encoded by the DNA, it is a protein.

PTS: 1 2. ANS: B

Genes are always present within a cell’s nucleus. However, they only produce the proper protein when they are activated, read, and transcribed.

PTS: 1 3. ANS: D

Transcription is the process of making a strand of RNA that is complementary to the DNA sequence that contains the gene for the protein needed. During DNA replication, both of the double strands of DNA within one cell are entirely copied, resulting in the total synthesis of two new complete strands. During protein synthesis, only the segment of DNA that contains the antisense strand that is complementary for the actual gene is involved in the process, not the entire genome. This means that only a segment of one DNA strand is read and transcribed into RNA. PTS: 1 4. ANS: A

The introns are not part of the gene and must be spliced out to form the mature messenger RNA that contains only the information encoded in the exons (expressed regions of a gene). In RNA, which is complementary to the DNA of the “sense strand,” thymine is replaced with uracil.

Therefore, response B is incorrect because it contains thymine. Response C is incorrect because it shows the segments corresponding to the introns and not the exons. Response D is incorrect because it shows retention of both the exons and the introns.

PTS: 1 5. ANS: B

A ribosome is a cytoplasmic adapter molecule containing a complex of proteins and some RNA that essentially decodes the mRNA and places the proper individual amino acid into the growing peptide chain during protein synthesis. It does not have anything to do with double-stranded DNA, nor does it perform any posttranslational modification. The transport molecules are the transfer RNAs (tRNAs), not the ribosomes.

PTS: 1 6. ANS: D

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Converting the early transcript of mRNA into mature mRNA requires splicing out the introns, which are the intervening sequences that are not part of the gene encoding for a specific protein.

Only the exons (expressed regions) of the initial transcript should remain in the mature messenger RNA ready for translation.

PTS: 1 7. ANS: D

DNA transcription, RNA transcription, and splicing out of introns occur in the nucleus to develop mature mRNA, which then moves into the cytoplasm for translation.

PTS: 1 8. ANS: C

Protein synthesis requires adequate amounts of all four bases (as nucleotides). If there are inadequate amounts of any one, all protein synthesis will be decreased or halted in that cell.

PTS: 1 9. ANS: A

Translation of mRNA into protein occurs in the cytoplasm, whereas transcription occurs in the nucleus. Transcription as a process would be unaffected by the lack of uracil; however, the transcript would have only thymine, not uracil. The methyl group of thymine would prevent the transcript from leaving the nucleus and entering the cytoplasm where translation and protein finishing can occur.

Thus, translation would be directly disrupted, and posttranslational modification would be indirectly disrupted. MicroRNA silencing is not part of this process.

PTS: 1 10. ANS: C

RNA is complementary to the antisense (template) strand of DNA. It also does not contain thymine (T). Instead, wherever an adenine (A) is present in the DNA, uracil (U) is placed in RNA. PTS: 1 11. ANS: A

A “TATA” box is a transcription start signal located upstream (5' to 3') of a specific gene. Anemia is a physiologic trigger that more beta globin should be synthesized. Polycythemia indicates an excess of red blood cells and no need for beta-globin production. Polyadenylation only finishes the transcription and does not signal the need to start it.

PTS: 1 12. ANS: C

Proteins are not in their final forms for active function when they are first synthesized and require posttranslational modification, the further processing of the newly translated primary protein structure into at least its secondary and tertiary structures to make it fully functional.

Secondary protein structure is a twisting of the primary structure as a result of the interaction of amino acids located near each other. Tertiary structure is the folding of the linear structure and occurs as a result of remote amino acids interacting with each other. Folding often creates a “pocket” within the protein that becomes an “active site,” able to interact with other structures or substances.

PTS: 1

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13. ANS: C

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The amino acid attachment site is the location that a specific amino acid can attach to and be carried by any one tRNA. Which amino acid attaches depends on the tRNA’s anticodon, which is the tRNA complementary code for an amino acid codon. Thus, for every RNA codon, there is a corresponding complementary anticodon on the tRNA that can attach and carry the correct amino acid. (Every single amino acid has its own specific tRNAs.) PTS: 1 14. ANS: C

Each amino acid is coded for by a triplet of bases in the DNA, which corresponds to the complementary triplet of bases composing the codon in RNA for each amino acid. Because each amino acid codon has three bases, the minimum number of bases needed in the mature messenger RNA for glucagon is 29 multiplied by 3, or 87.

PTS: 1 15. ANS: B

Every active protein has a specific amount of the amino acids and a unique sequence in which they are connected together. The exact sequence is critical for protein function. It is possible for two separate proteins to have the same total number of amino acids and perhaps even the same numbers of individual amino acids (so response A is incorrect). However, the sequencing order of the amino acids is what makes one protein different in structure and function from another protein. The exons are the actual directions for the sequence of amino acids. Deleting these would not result in a functional protein. Transcription always occurs before, not after, translation in the process of protein synthesis.

PTS: 1 16. ANS: D

Codons are the RNA code for specific amino acids within a protein. When these are read properly, the correct amino acid is placed properly into the growing peptide chain to lead to the formation of a functional protein.

PTS: 1 17. ANS: A

RNA does not contain the pyrimidine base thymine. The base uracil is used in place of thymine because it is a pyrimidine base with a structure that does not contain the methyl group (CH3) that thymine has. This difference between thymine and uracil is important because molecules in the nucleus that contain a methyl group remain trapped inside the nucleus. Because the remaining phases of protein synthesis occur outside the nucleus, the newly transcribed RNA must be able to exit the nucleus.

PTS: 1 18. ANS: D

The addition of a poly-A tail to the newly transcribed RNA, known as polyadenylation, results in a segment of RNA that contains mostly adenine and is not translated into part of the protein.

Thus, it serves as a signal to stop translation.

PTS: 1 19. ANS: A

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Once in the cytoplasm, mRNA molecules have a very short life span, only seconds, before they are degraded by enzymes known as ribonucleases (RNases). This rapid degradation of mRNA is important in preventing an inadvertent overproduction of specific proteins. The idea is to make just enough active protein as is needed at that time and no extra. This makes protein synthesis less wasteful and more efficient. It also prevents too much of a specific protein from being present and exerting effects that are not needed.

PTS: 1 20. ANS: D

Introns are the sectional parts of DNA within a gene-coding region that do not belong to the gene-coding sequence of the protein being synthesized. However, because these introns are in gene-coding regions, they are parts of another gene. In that other gene, they would be considered exons for that gene. Thus, they are not part of the desert DNA and have no role in the synthesis of the gene product for which they are introns.

PTS: 1 21. ANS: B

Because an amino acid is encoded in the DNA by a “triplet” of bases, deletion of any number of bases that is not a multiple of three will alter the reading frame and result in a frameshift mutation. Although deletion of “triplet” bases can result in a change in some areas of the amino acid sequence and have an influence on protein function, the essential reading frame is not disrupted.

PTS: 1 22. ANS: C

The most common form of a normal functional gene is known as the wild type. Thus, a person with the wild-type form of the HD gene has a functional gene with no increased risk for developing the disease.

PTS: 1 23. ANS: C

A nonsense point mutation results in an inappropriate placement of a stop signal, which has a negative effect on protein function. This type of mutation prevents the completion of a protein.

The protein may not be synthesized at all if the stop signal is present early in the reading sequence. If it is present later in the sequence, protein synthesis stops prematurely and results in a short or truncated protein that usually has little if any function.

PTS: 1 24. ANS: D

Frameshift mutations are disruptions of the DNA reading frame (not the chromosome) as a result of having a whole base or group of bases added or deleted. They can occur in somatic cells or germline cells. When this type of mutation occurs in gene-coding regions, it always disrupts the reading frame from the start of the mutation to the end of the gene. The result is complete alteration of amino acid position and prevention of synthesis of a functional protein. A normal protein cannot be made from a gene with a frameshift mutation. Although mutations may accumulate over a lifetime, frameshift mutations do not occur more often than point mutations as a person ages.

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PTS: 1

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25. ANS: A

Mutations of any type that occur in noncoding regions are responsible for making one person’s DNA different from and identifiable from another person’s DNA. Even identical twins (monozygotic twins) do not have absolutely identical DNA by the time they are born, although they probably did when the embryo first split into two embryos. By the time identical twins are born, they usually have at least 100 base pairs different from each other in the noncoding regions.

As they live their lives, each twin continues to accumulate more and different mutations so that as they age, these identical twins become less identical in their DNA. PTS: 1 26. ANS: D

Point mutations are substitutions of one base for another and can occur in DNA or RNA. This type of change does not result in an extra base or a lost base, just a substitution. This type of base change is known as a single-nucleotide polymorphism (SNP). Frameshift mutations are deletions or insertions of DNA bases, not one-for-one substitutions. SNPs do not interconvert introns and exons.

PTS: 1 27. ANS: B

Everyone experiences some mutational events as a result of spontaneous DNA replication error or exposure to mutagens or carcinogens in the environment. Many of these mutational events are correctly repaired and have no lasting consequences. However, when they remain unrepaired and occur in a gene-coding region for cell growth regulation, they can have permanent consequences for the individual, including a greater risk for cancer development.

PTS: 1 28. ANS: B

Somatic cell mutations occur only in ordinary body cells, not in germline cells (eggs or sperm).

Thus, somatic mutations cannot be passed on to one’s children. The presence of somatic mutations is a major cause of sporadic cancer in a person, but this predisposition cannot be inherited by his or her children.

PTS: 1 29. ANS: C

Tertiary structure is the folding of the linear structure and occurs as a result of remote amino acids interacting with each other. These interactions, including bod formation, allow parts of the linear structure to draw closer together in some areas and have greater distances in other areas.

These shape-changing factors influence a protein’s activity. The poly-A tail only finishes the transcription and is not translated. Additional proteins in specific association are a protein’s quaternary structure. The specific amino acids in close proximity to each other in a peptide help for the protein’s secondary structure.

PTS: 1 30. ANS: C

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It is most likely that the mutational events in a noncoding region will not interfere with correct reading of any genes. These changes in the noncoding regions can make Jack’s DNA more identifiable. With Jill having a frameshift mutation in the gene-coding region, it is likely that she will not be able to produce functional insulin and will have type 1 diabetes mellitus, requiring insulin replacement. Because this large mutational event (all the pancreatic beta cells) is occurring only in somatic cells, Jill cannot pass this problem on to her children.

PTS: 1 31. ANS: A

MicroRNA is a small, noncoding piece of RNA that regulates gene expression at the RNA level by binding to parts of targeted mRNA molecules, making them partially double stranded, which cannot be translated. This action “silences” the translation ability of selected mRNA molecules and increases the rate at which they are degraded. This experimental drug has the potential to increase insulin production by allowing insulin-coding mRNA to be present longer and not have its translation inhibited by being double stranded. MicroRNA has no effect on a protein once it has been produced. Thus, the degradation of formed insulin is not changed.

PTS: 1

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Chapter 3: Genetic Influence on Cell Division, Differentiation, and Gametogenesis Multiple Choice Identify the choice that best completes the statement or answers the question.

1. In what way is hypertrophic tissue growth more advantageous than hyperplastic tissue growth?

A. There is no limit to how large a tissue or organ can become.

B. It proceeds at the same rate throughout a person’s life span.

C. Less energy is required for hypertrophic growth.

D. Differentiated functions change with aging.

2. In which phases of the cell cycle is the normal cell tetraploid (4N)?

A. G1 and G2 B. G1 and S C. S and G2 D. G2 and M 3. How is apoptosis related to physiologic homeostasis?

A. The process prevents germline mutations.

B. The efficiency of organ/tissue functions is increased.

C. Replacement with scar tissue occurs more rapidly after cell damage.

D. Cells that are able to undergo apoptosis bypass restriction point controls

for mitosis.

4. Which normal cell characteristic is represented by the production of insulin in the beta cells of

the pancreas?

A. Performance of a differentiated function B. Ability to undergo apoptosis on schedule C. Tight regulation of cell division D. Conservation of energy 5. What is the consequence for a tissue/cell that no longer produces any cell adhesion molecules?

A. The production of three daughter cells with mitosis instead of just two B. Failure to perform any differentiated functions C. Conversion from euploidy to aneuploidy D. Migration of cells into other tissues 6. How do transcription factors influence cell division?

A. They directly transmit external signals to the cell’s nucleus.

B. Transcription factors regulate the expression of genes involved in cell division.

C. Loss or inactivation of transcription factors degrades suppressor gene

protein products.

D. Transcription factors are enzymes that activate promitotic substances by

adding a phosphate group to the chemical structure.

7. How does apoptosis contribute to healthy organ and whole-body function?

A. Maintains telomeric DNA B. Allows cells to differentiate rather than to undergo mitosis

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C. Removes old or damaged cells from an organ population D. Activates oncogenes when cells are damaged or necrotic 8. What would be the expected response to a skin injury if the involved tissue had lost the normal

cell characteristic of contact inhibition?

A. Failure of the wound to close B. Excessive growth of replacement tissue C. Replacement with scar tissue rather than skin D. Displacement of skin cells into other body tissues

9. What is the function of a suppressor gene product?

A. To ensure cell division occurs only when it is needed B. To suppress the loss of differentiated functions with aging C. To ensure the precise delivery of chromosomes to each new daughter cell D. To prevent the formation of a large nuclear-to-cytoplasmic ratio during the G0 state 10. Normal cells spend most of their life spans in which phase?

A. G0 B. G1 C. S D. M 11. How do cyclins influence the process of cell division?

A. Suppressing oncogene products and inhibiting movement through the cell cycle B. Generating transcription factors and promoting differentiated functions C. Promoting apoptosis and allowing programmed cellular “suicide” D. Opposing suppressor gene products and promoting cell division 12. What is the general purpose of tyrosine kinase enzymes?

A. Activating cyclins by phosphorylation B. Increasing the amount of sodium-driven transcription factor activity C. Suppressing the activation of oncogene expression D. Uncovering growth factor receptor sites on plasma membranes 13. Which event occurs during mitosis?

A. Homologous chromosomes synapse and then cross over.

B. The number of chromosomes decreases from diploid to haploid.

C. Daughter cells are produced that are genetically identical to the parent cell.

D. DNA density decreases, and the 46 separate pieces assemble into one linear strand.

14. Which cell feature is common to normal human differentiated cells and to early embryonic

human cells?

A. Growth by hypertrophy B. Contact inhibition C. Tight adhesion D. Euploidy

15. Which feature or characteristic of early embryonic cells is unique in comparison with normal

differentiated cells?

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A. Diploidy B. Pluripotency C. Controlled cell division D. Mitosis resulting in four new daughter cells 16. Which event characterizes embryonic commitment?

A. Meiotic cell division B. Selective loss of genes C. Increased suppressor gene activity D. Progressive increase in nuclear size 17. Which mechanism allows cell differentiation in embryonic tissue?

A. Increased expression of oncogenes B. Decreased expression of suppressor genes C. Selected expression of individual structural genes D. Enhanced expression of promitotic transcription factors 18. How many different genotypes are possible in any single mature spermatocyte of a man who is

heterozygous at a single gene locus for a specific trait?

A. One B. Two C. Four D. Eight 19. Which stage of cell division is present in mitosis but is missing in meiosis?

A. G1 B. S C. G2 D. M 20. What is the expected result when two homologous chromosomes fail to separate during meiosis I

of spermatogenesis?

A. Fertilization of this gamete may result in a zygote that is 48, XXYY. B. One gamete will have two of these chromosomes, and one will have neither of these chromosomes.

C. The risk for development of polygenic disorders is increased with fertilization of the polar body associated with this mature gamete.

D. Increased genetic diversity is possible with fertilization of any of these four gametes because of an increased number of possible gene alleles.

21. During which phase of gametogenesis is the process of crossing over more likely to occur for

either spermatocytes or oocytes?

A. Metaphase I B. Interphase I C. Anaphase I D. Prophase I 22. Which process is unique to spermatogenesis?

A. Synapsis during the zygotene stage

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B. Equal distribution of cytoplasm during meiosis I C. Elongation of chromatids during the diplotene stage D. Degeneration of germ cells between fetal life and the onset of puberty 23. Which cell in the process of oogenesis has the most chromosomes?

A. Mature ovum B. Primary oocyte C. Secondary oocyte D. Polar body 24. What is the usual outcome of “crossing over” during meiosis I for both spermatogenesis

and oogenesis?

A. Unequal division of DNA between two new daughter cells during nucleokinesis B. Mixing of maternal and paternal genes within one chromosome pair C. Unequal distribution of cytoplasmic material D. More rapid progression to meiosis II 25. Which cell division process sequences are normal for meiosis for gametogenesis?

A. Two rounds of DNA synthesis each followed by a separate round of meiotic cell

division B. Two rounds of DNA synthesis followed by two progressive rounds of meiotic cell division C. A single round of DNA synthesis followed by two separate rounds of meiotic cell division D. A single round of DNA synthesis first preceded by one round of meiotic cell division and then followed by a final round of meiotic cell division 26. Which response is the immediate and direct result of fertilization?

A. Gamete chromosome reduction to the haploid number B. Rapid proliferation of acrosomal and coronal cells C. Primary sex determination of the zygote D. Nuclear condensation 27. What is the consequence of synapsis and crossing over?

A. Pure segregation of alleles along the metaphase plate B. Tetraploidy, in which there are four copies of each chromatid C. Diploidy of chromosome number and haploidy for DNA

content

D. Random recombination of genetic material between paternal and maternal

chromatids 28. Why is meiosis II for both spermatogenesis and oogenesis called an “equatorial division?”

A. The amount of cytoplasm in the secondary sex cells is evenly divided among the

final cells.

B. The actual number of chromosomes within the resulting cells is the same as before this division.

C. Crossing over stops, and the bivalent chromosomes coil and condense in preparation for segregation.

D. The division of both the two secondary spermatocytes and the two secondary

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oocytes results in a total of eight gametes.

29. How many mature ovum result from the complete oogenesis of one oogonium?

A. One B. Two C. Three D. Four 30. Meiosis II of oocytes is completed at which developmental period?

A. At the ninth prenatal week B. During puberty C. At ovulation D. At fertilization 31. Why is normal fertilization of one mature ovum usually performed by only one mature sperm

even though hundreds of millions of sperm are present in the seminal fluid of one ejaculation?

A. The corona radiata causes the tail of the other sperm to fall off, decreasing their motility.

B. At the time of first penetration, the pH around the ovum changes, and the enzymes of the other sperm are inactivated.

C. Fusion of the two nuclei forms a tight nuclear membrane that prevents the nuclei of other sperm from entering.

D. After being penetrated by one sperm, the ovum’s membrane changes electrically and prevents other sperm from entering.

32. Why is fertilization of a polar body unlikely to lead to normal embryonic and fetal development?

A. The resulting zygote would be 4N instead of 2N. B. The lack of cytoplasm would inhibit cellular reproduction.

C. The polar body has spent too long of a time trapped in meiosis I. D. The resulting zygote would be smaller, which increases the risk for apoptosis.

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Chapter 3: Genetic Influence on Cell Division, Differentiation, and Gametogenesis Answer Section MULTIPLE CHOICE 1. ANS: C

More energy is used in the replacement process of tissue growth by hyperplasia than by hypertrophy. Even with hypertrophy, there is a limit to tissue or organ size. Most periods of hypertrophy for normal cells occur during periods of development, such as childhood and adolescence. Although skeletal muscle can hypertrophy with exercise even in older adulthood, it does not proceed at a uniform rate throughout life, nor does the actual differentiated function of the tissue change.

PTS: 1 2. ANS: C

The cell cannot be tetraploid until DNA synthesis occurs to double the DNA content. In M phase, the chromosomes split and are sent to both daughter cells to make them diploid. Therefore the two phases in which the cell is tetraploid are S phase and G2.

PTS: 1 3. ANS: B

In tissues and organs that are composed of cells capable of undergoing apoptosis (programmed cell death), older or poorly functional cells are removed, making room for new cells generated by mitosis so that the tissue/organ continues to be populated throughout life by optimally functional cells.

Apoptosis has no role in preventing or altering germline mutations. Replacing dead or damaged cells with scar tissue reduces organ function and does not contribute to physiologic homeostasis.

Cells that undergo apoptosis must respond to the same mitotic restrictions as cells that do not undergo apoptosis.

PTS: 1 4. ANS: A

Although the gene for insulin is present in all body cells, it is only expressed in the beta cells of the pancreas. Thus, the gene product insulin is an example of a specific differentiated function performed by one cell type. Energy is used by the beta cells in the performance of this differentiated function. The process does not represent either regulation of cell division or the capability of apoptosis.

PTS: 1 5. ANS: D

Most normal cells have cell adhesion molecules, which are cell surface proteins that allow normal cells of the same type to adhere tightly together. As a result, they are not migratory.

Without cell adhesion molecules, tissues and organs would have a looser construction in which cells could break off and migrate into other body areas. Cell adhesion molecules have little effect on the performance of differentiated functions and do not regulate the chromosome number of a cell. Normal cell division does not result in the formation of three new daughter cells regardless of whether cell adhesion molecules are present or are not present.

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PTS: 1

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6. ANS: B

Transcription factors are proteins that enter a cell nucleus and regulate transcription for a specific gene or set of genes. They are intermediaries and work indirectly after receiving external information to transmit signals to the nucleus. They are not degradative enzymes. Tyrosine kinases are enzymes that activate promitotic substances by adding a phosphate group to the chemical structure, not transcription factors.

PTS: 1 7. ANS: C

Apoptosis is programmed cell death or cellular suicide of tissue cells. It allows poorly functional cells to be removed from organs and tissues, making room for new cells to generate by mitosis so that the tissue or organ continues to be populated throughout life by optimally functional cells without increasing in size. If apoptosis did not occur, some tissues and organs would contain too few functional cells to perform their work efficiently and effectively. Apoptosis does not maintain telomeric DNA, activate oncogenes, or stimulate differentiation.

PTS: 1 8. ANS: B

Once a normal cell is completely surrounded by other cells and its membrane is contacted directly on all surface areas with the membranes of other cells, it no longer undergoes mitosis. This characteristic is known as “contact inhibition of cell growth.” The purpose of this characteristic is to prevent inappropriate tissue overgrowth.

PTS: 1 9. ANS: A

Normal cells have their growth regulated by a balance between products produced by oncogenes, which promote entering and completing the cell cycle, with those of products produced by suppressor genes, which restrict or inhibit entering and moving through the cell cycle. Thus, suppressor gene products inhibit all aspects of mitosis and also trigger apoptosis. As a result, cell division occurs only when it is needed, and environmental conditions can support more cells.

PTS: 1 10. ANS: A

Cells not actively reproducing (undergoing mitosis) are outside of the cell cycle in G0, the reproductive resting state, and continue to perform all their usual differentiated functions. All cells spend more time in this state than actively dividing. G1, S, and M phases are all part of the reproductive cycle of cell division.

PTS: 1 11. ANS: D

Cyclins are a group of promitotic proteins produced by specific oncogenes that, upon activation, propel the cell forward through all phases of the reproduction cycle. Normally, the oncogene expression of cyclins is carefully regulated by suppressor gene products, not the other way around. They respond to transcription factors rather than generate them. Cyclins actually work in opposition to apoptosis.

PTS: 1 12. ANS: A

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Cyclin activation requires the attachment of a phosphorous molecule to the cyclin structure, a process known as phosphorylation, which is performed by a variety of tyrosine kinases (TKs).

They do not directly affect sodium-driven transcription or suppress oncogene expression. They are present inside cells and cannot uncover surface receptor sites.

PTS: 1 13. ANS: C

The outcome of mitosis is the formation of two daughter cells that are genetically identical to the parent cell that initiated cell division. Selections A and B are events of meiosis only and do not occur as part of mitosis. Selection D is incorrect in that the 46 separate strands of DNA never assemble into one linear strand.

PTS: 1 14. ANS: D

Both normal human differentiated cells and normal early embryonic cells have the correct number of chromosomes, 23 pairs. Early embryonic cells do not grow by hypertrophy, nor do they demonstrate contact inhibition of cell growth. Only differentiated cells are tightly adherent.

PTS: 1 15. ANS: B

Early embryonic cells are pluripotent rather than differentiated. They have the potential to become any body cell. Normal differentiated cells have committed to one pathway and do not have the potential to become other cell types. Normal differentiated cells and early embryonic cells are euploid and diploid. Normal mitosis of embryonic cells and of differentiated cells never results in four new daughter cells. Although early embryonic cells reenter the cell cycle rapidly and spend very little time in the reproductive resting state of G0, cell division is still considered controlled.

PTS: 1 16. ANS: C

Commitment involves adjusting the activity of the promitotic oncogenes and the genes that regulate differentiation. Suppressor gene activity increases so that greater control and limits are placed on oncogenes, usually slowing cell division somewhat. In addition, whatever genes are important for structure and function within specific organs are selectively expressed, but no genes are lost from any cell type. Meiotic cell division occurs later in fetal life for females and not until puberty for males. It is not a function of commitment. With commitment and change in cell division, the nuclear size decreases.

PTS: 1 17. ANS: C

Normal development requires that early embryonic cells lose pluripotency and commit to a specific differentiation. Commitment involves adjusting the activity of the promitotic oncogenes and the genes that regulate differentiation. Suppressor gene activity increases so that greater control and limits are placed on oncogenes, usually slowing cell division somewhat. In addition, whatever genes are important for structure and function within specific organs are selectively expressed.

PTS: 1 18. ANS: A

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The mature spermatocyte is haploid, with only one chromosome of each pair present. It should have only one gene allele. Even though the person is heterozygous for a single gene trait and his diploid cells have two alleles, each mature spermatocyte will have only one. In about half of the mature spermatocytes, one specific allele of the two is present. In the rest of the mature spermatocytes, the other specific allele is present.

PTS: 1 19. ANS: C

For meiosis, there is no G2 phase. Cells begin M phase, which is prolonged, immediately after leaving S phase.

PTS: 1 20. ANS: B

Failure of a chromosome to separate properly during meiosis is nondisjunction. The result of nondisjunction is that one cell will have two copies of the chromosome, and the other cell will not have a copy of this chromosome. Depending on which chromosome is missing or present in two copies, viable embryos may be produced from fertilization, either as monosomies or trisomies.

This process does not result in a zygote with 48 chromosomes, nor will genetic diversity be increased.

Polar bodies do not form in spermatogenesis.

PTS: 1 21. ANS: D

Crossing over is the lengthening and touching of the chromatids that allows for pieces of sister chromatids to be exchanged. For both spermatogenesis and oogenesis, the process occurs during prophase I, although the timing differs between the two processes.

PTS: 1 22. ANS: B

In oogenesis the first meiotic cell division results in the unequal distribution of cytoplasmic fluid and all organelles into the secondary oocyte. As a result, the first polar body contains very little of these elements. By the end of meiosis II, only the mature ovum has cytoplasm. In spermatogenesis, the cytoplasm and all its elements are evenly distributed to the secondary spermatocytes (and eventually to the four spermatids). Synapsis occurs during the zygotene stage of both spermatogenesis and oogenesis. Chromatid elongation occurs in the leptotene stage.

Degeneration of germ cells between fetal life and the onset of puberty occurs only in oogenesis.

PTS: 1 23. ANS: B

The primary oocyte first starts oogenesis by entering the cell cycle and proceeding through S phase with DNA replication, which results in tetraploidy or a 4N state. The mature ovum and polar bodies are haploid, containing only 1N of chromosomes. The secondary oocyte has undergone meiosis I and is 2N. PTS: 1 24. ANS: B

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In gametogenesis, crossing over occurs when the four chromatids of each chromosome pair are lined up lengthwise and touch; the exchanges of chromosome material occur through breaks and rearrangements. The exchanges occur not just between chromatids from one parent but also among the four chromatids for both the maternal and paternal chromosomes of the pair. The result is a huge “shuffling” of genetic material, so at the end of the pachytene stage, the two chromosomes (with two chromatids each) are now combinations of maternal and paternal genes, rather than one purely maternal-derived chromosome and one purely paternal-derived chromosome.

PTS: 1 25. ANS: C

Cell division of meiosis for gametogenesis is a special type of cell division in which the chromosome number per cell is reduced to half and occurs only in germ cells. It involves only one episode of DNA synthesis that is followed, over time, by two separate rounds of meiotic cell divisions.

PTS: 1 26. ANS: C

At fertilization, the union of one mature haploid sperm with one mature haploid ovum occurs to form a diploid zygote. One sex chromosome is provided by the spermatocyte, and one is provided by the mature ovum, determining the primary or “genetic” sex of the new zygote. The nucleus increases in size and does not condense at this time. The gametes are haploid long before fertilization occurs. Changes in the acrosome of the spermatocyte and coronal cells of the ovum also occur before fertilization, not after.

PTS: 1 27. ANS: D

Synapsis allows the homologous chromosomes with four chromatids to remain in contact down the lengths of the chromatids for both the maternal and paternal chromosomes of the tetraploid pair in the early gamete. This results in a huge but usually even “shuffling” of genetic material so that at the end of the pachytene stage, the two chromosomes (with two chromatids each) are now combinations of maternal and paternal genes, rather than one pure maternally derived chromosome and one pure paternally derived chromosome. This allows for any one person to receive bits and pieces of genetic material in combination from many parental ancestors.

PTS: 1 28. ANS: B

For both ova and sperm, meiosis II is a relatively rapid process. This division is an equational division because the number of chromosomes remains the same (23), although the DNA content was diploid before the division and haploid after the division.

PTS: 1 29. ANS: A

Complete oogenesis is the process of forming a mature ovum from a precursor female germ cell known as oogonium. Unlike spermatogenesis, in which one precursor male germ cell normally produces four spermatocytes, only one ovum is produced by this process, with up to three polar bodies also produced.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 30. ANS: A

Completion of meiosis I of the primary oocyte into a secondary oocyte and a polar body does not happen until just before ovulation. Meiosis II of the secondary oocyte is completed only if fertilization takes place.

PTS: 1 31. ANS: D

When the sperm binds with and then penetrates the plasma membrane of the ovum, several different processes occur. First, the ovum’s plasma membrane changes its electrical charge, preventing any other sperm from entering. Only the fertilizing sperm’s tail and midsection drop off and do not enter the ovum; the other sperm remain motile but cannot penetrate the ovum’s plasma membrane. It is not the nuclear membrane that prevents two (dispermy) or more sperm from fertilizing a single ovum. Although the pH of fluid in the female reproductive tract can inhibit acrosomal enzyme activity, this is not what prevents dispermic fertilization.

PTS: 1 32. ANS: B

Polar bodies are haploid (1N) and, in theory, could be fertilized to form a zygote. However, because both mature sperm and polar bodies are almost completely deficient in cytoplasm and mitochondria, any fertilized polar body would not have the energy or the intracellular machinery to direct normal development.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 4: Patterns of Inheritance Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which factor allows inheritance patterns for a specific trait or health problem to be traced from

one family generation to another?

A. Pleiotropy B. Allele segregation C. Regression to the mean D. Gene–environment interaction 2. If there are 10 possible alleles for the single-gene trait of nose shape, how many alleles can a

person with euploid chromosomes inherit from his or her biological parents?

A. 1 B. 2 C. 4 D. 5 3. An infant with type B blood is born to parents who both have type O blood. What genetic action

or phenomenon is responsible for this unusual blood type expression?

A. Decreased penetrance of a dominant trait B. Variable expressivity of a recessive trait C. Inactivation of both parents’ X chromosomes D. Failure of one parent’s blood-type gene alleles to segregate 4. Is it possible for two parents with achondroplasia to have a child who is of normal stature?

A. Yes, because the disorder is autosomal dominant, and if both parents

are heterozygous, the child could inherit two normal stature gene alleles.

B. Yes, if the parent who is homozygous for the gene mutation demonstrates variable expressivity of the health problem.

C. No, because the disorder is autosomal recessive, and the child can only inherit two mutated gene alleles for the health problem.

D. No, because homozygosity for this health problem is lethal.

5. What is the risk for a person to inherit an autosomal-dominant genetic disease–causing allele

from a parent who is heterozygous if the disorder has a penetrance factor of 75%?

A. 100% B. 75% C. 50% D. 25% 6. Which statement reflects the criterion for autosomal-dominant transmission of single-gene traits?

A. The risk for a person who is homozygous for the trait to transmit the trait to his

or her children is 100% with each pregnancy.

B. The trait often remains unexpressed within a kindred for many generations

until a change in environment promotes its expression.

C. Males in a kindred are more likely to express the trait when the mother has the

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

trait, and females in a kindred are more likely to express the trait when the father has the trait.

D. Females in a kindred are more likely to express the trait when the mother has the trait, and males in a kindred are more likely to express the trait when the father has the trait.

7. In which situation are phenotype and genotype always the same?

A. X-linked-recessive traits B. X-linked-dominant traits C. Autosomal-recessive traits D. Autosomal-dominant traits 8. A girl of normal stature is born to two parents with achondroplasia who have very short stature,

especially disproportionately short arms and legs. What is the probability (by Punnett square analysis) that any pregnancy this girl eventually has will result in the birth of an infant with achondroplasia if her partner also has normal stature?

A. Three out of four (75%) B. Two out of four (50%) C. One out of four (25%) D. Zero out of four (0%) 9. Which statement or factor is a criterion for autosomal-recessive transmission of single-gene

traits?

A. About 25% of the members of a large kindred with an autosomal-recessive trait will express the trait.

B. There is no carrier status; if the allele for the trait is present, it is expressed, although the degree of expression can be variable.

C. Individuals who are heterozygous for an autosomal-recessive trait have minimal risk for transmitting the allele to their offspring.

D. The degree of expression of an autosomal-recessive trait or disorder in a homozygous individual is directly related to the penetrance of the trait.

10. Which person is an obligate carrier of an autosomal-recessive single-gene trait or disorder

without expressing the trait or disorder?

A. The son of a man with classic hemophilia B. The daughter of a woman with Marfan syndrome C. The son of a man who expresses a widow’s peak hairline D. The daughter of a woman who expresses attached earlobes 11. With which type of inheritance pattern does the trait or disorder usually first appear among

siblings rather than in parents of affected children?

A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. Codominant 12. Which type of genetic transmission promotes the continued existence of genetic mutations in

single genes?

A. Autosomal dominant

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B. Autosomal recessive C. Codominant D. Sex-linked 13. On a five-generation pedigree, which feature distinguishes an X-linked-dominant disorder in

which males and females are equally affected from an autosomal-recessive pattern of inheritance?

A. The disorder is not apparent until adulthood.

B. Unaffected individuals do not have affected children.

C. There are no instances of an affected father transmitting the disorder to his son.

D. Unaffected parents (who do not express the disorder) do have affected children.

14. A woman whose father does not have hemophilia gives birth to a son with classic hemophilia.

This woman is found to have only 15% of the normal amount of clotting factor VIII (the clotting factor males with hemophilia are missing) and does have abnormal blood-clotting issues. Which phenomenon or factor is most likely responsible for her abnormal clotting factor expression?

A. The X chromosome most commonly inactivated in her bone marrow is paternally derived.

B. The X chromosome most commonly inactivated in her bone marrow is maternally derived.

C. This woman is demonstrating reduced penetrance of an X-linked-recessive disorder.

D. This woman is demonstrating full penetrance of an X-linked-recessive disorder.

15. Which statement regarding inheritance of an autosomal-dominant gene allele with known

variability in expressivity is true?

A. A person with low expressivity of the trait has a higher probability of having a child who does not express the trait at all.

B. A person with high expressivity of the trait has a greater risk of having a child who expresses the trait to an even greater degree.

C. The degree of expressivity of a given autosomal-dominant trait with known variability cannot be predicted by analyzing parental expression.

D. The degree of expressivity of a given autosomal-dominant trait with known variability is greater when the transmitting parent is the same sex as the child.

16. A man whose parents both have brown hair claims that his red beard was inherited from his

maternal uncle. Why is this claim incorrect?

A. The Y chromosome has no role in scalp, facial, or body hair color.

B. Genetic traits are transmitted in only a direct vertical direction.

C. Females cannot transmit facial hair color to their sons.

D. Hair color is a polygenic trait, not a single-gene trait.

17. Mating of a yellow male parakeet with a blue female parakeet resulted in 18 offspring. Four

(two males; two females) had blue feathers, and 14 (eight females and six males) had yellow feathers. What allelic combinations and mode of inheritance can you determine from the evidence of the feather colors expressed?

A. Yellow allele dominant; blue allele recessive; autosomal B. Yellow allele recessive; blue allele dominant; autosomal

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

C. Yellow allele dominant; blue allele recessive; sex-linked

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

D. Yellow allele recessive; blue allele dominant; sex-linked 18. Mating of a female rabbit with a long tongue with a male rabbit that had a short tongue resulted in

12 offspring. Two had long tongues, two had short tongues, and eight had medium-length tongues.

Which allele combinations explains the tongue lengths of the offspring?

A. Short tongue dominant; long tongue recessive B. Short tongue recessive; long tongue dominant C. Short tongue dominant; long tongue dominant D. Short tongue dominant; long tongue X-linked recessive 19. Which trait or disorder has both an autosomal-dominant form and an autosomal-recessive form?

A. Widow’s peak B. Diabetes mellitus type 2 C. Polycystic kidney disease D. Taste sensitivity to phenylthiocarbamate (PTC) 20. What is the most likely explanation for the expression of responses related to pleiotropy?

A. The same gene may have more than one chromosome locus.

B. The protein coded for by the single gene is an important component of more than

one tissue.

C. The silencing of a gene allele inherited from one parent is expressed differently depending on which parent contributed the silenced allele.

D. The same disorder may have an autosomal-dominant inheritance pattern, an autosomal-recessive inheritance pattern, and an X-linked-recessive inheritance pattern.

21. Which statement is a criterion for an autosomal-dominant pattern of inheritance of a specific trait

or characteristic that is highly penetrant?

A. Carriers for the trait may express it but do not necessarily express the trait.

B. Unaffected family members do not transmit the trait to their children.

C. Genotypes of individuals expressing the trait must be homozygous.

D. The trait appears only among male offspring of female carriers.

22. Which statement or condition best reflects multifactorial inheritance?

A. A mutation in a single gene results in the expression of problems in a variety of

tissues and organs.

B. The susceptibility to a problem is an inherited trait, but development of the problem is related to environmental conditions.

C. A mutated gene is inherited, but the results of expression of that gene are not evident until middle or late adulthood.

D. Several genes are responsible for the mechanism of hearing, and a mutation in any one of them results in hearing impairment.

23. What is the expected outcome of any pregnancy or child produced by two parents who are

each heterozygous for the Huntington disease allele?

A. The couple cannot produce children who are unaffected.

B. Homozygous offspring will express the disease in childhood instead of as an adult.

C. A child who is homozygous for the mutated alleles is not likely to transmit the

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

disorder to his or her children.

D. Because of the possibility of incomplete or low penetrance, an unaffected offspring could have an affected child.

24. A baby boy is born with six toes (polydactyly) on his right foot, a trait that neither parent has but

one that the paternal grandfather has. What is the best explanation for this occurrence?

A. The baby’s grandfather is really his father.

B. The baby inherited a recessive allele for this trait from each parent.

C. The father’s phenotype results from incomplete penetrance of a dominant allele.

D. The baby’s phenotype demonstrates a higher level of expressivity than his father’s phenotype.

25. What can be inferred about a genetic disorder that, when it occurs in monozygotic twins, affects

one twin 80% of the time and affects both twins only 20% of the time?

A. Expression of homologous genes is influenced by the gender of the parent who contributed them.

B. Nongenetic factors can influence the expression of identical alleles.

C. The mutation occurred in a somatic cell rather than a germ cell.

D. Mutation repair is incompletely penetrant.

26. A woman (Adele) has three daughters (Barbara, Brenda, Beverly) with a man (Adam) who is

color-blind. Barbara has a daughter (Carol) and a son (Cliff) who have normal color vision.

Brenda has no children. Beverly has two sons (Carl, Charlie) who are both color-blind. To which genetic generation do Carl and Charlie belong?

A. P1 B. P2 C. F1 D. F2 27. What is the expected expression of a monogenic trait that has its locus on the Y chromosome?

A. Females never inherit the trait.

B. Females inheriting the trait will be carriers.

C. Males inherit the trait in a dominant pattern of expression; females inherit the

trait in a recessive pattern of expression.

D. Females inherit the trait in a dominant pattern of expression; males inherit the trait in a recessive pattern of expression.

28. Which genetic disorder has both an autosomal-dominant and an autosomal-recessive form?

A. Syndactyly B. Phenylketonuria C. Long QT syndrome D. Retinitis pigmentosa 29. Which situation most closely represents an example of “regression to the mean”?

A. Two hearing-impaired parents produce a child who has normal hearing.

B. A 40-year-old man whose father developed type 2 diabetes mellitus at age 50 is

diagnosed with the disorder.

C. The three children of a mother who has an intelligence quotient (IQ) of 170 all

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

have IQs in the 110 to 120 range.

D. A child whose biologic parents are thin is adopted by a family whose members are obese and eventually becomes overweight.

30. Why are dizygotic twins less likely to show concordance for a monogenic trait than

monozygotic twins?

A. Dizygotic twins share fewer allele sequences in common than monozygotic twins.

B. It is possible for dizygotic twins to have different fathers, and monozygotic twins always have the same father.

C. Gene expression in monozygotic twins is less influenced by environmental factors than that of dizygotic twins.

D. Because of their identical appearance, monozygotic twins are more likely to be raised together and share a common environment than are dizygotic twins.

31. When calculating the recurrence risk for a complex, possibly polygenic disorder with a couple

who already have one child with the disorder, which additional factor has the greatest impact on increasing this risk?

A. The child with the disorder is male, and the overall incidence of the disorder is twice as high in males than females.

B. Both the mother and the father are heavily exposed to a variety of chemicals in their workplaces.

C. The father has a nephew with a mild form of the same disorder.

D. The mother has a brother who also has the same disorder.

32. Which common health problem has the highest heritability estimate?

A. Hypertension B. Peptic ulcer disease C. Congenital heart disease D. Schizophrenic behavior 33. Which statement about the X chromosome is true?

A. Most of the X chromosome genes encode proteins that have no role in female

sexual development.

B. Heterozygous females are more severely affected by X-linked-dominant

disorders than heterozygous males.

C. When dominant alleles are present on the X chromosome, they are expressed in

a recessive manner in the hemizygous state.

D. When males have the unusual condition of two Y chromosomes and are missing

the X chromosome, all physiologic function remains normal.

34. If a man with classic hemophilia (X-linked recessive) has children with a woman who is a carrier

for the disorder, what is the expected risk pattern?

A. All sons will be unaffected; all daughters will be carriers.

B. All sons will be carriers; all daughters will be affected.

C. All sons have a 50% risk of the disorder; daughters will all either be affected or carriers.

D. All sons will be carriers; daughters have a 50% chance of being a carrier and a 50% chance of being unaffected.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

35. What is the significance of a shift to the left in the threshold value of liability in a threshold

of liability model for the expression of a complex health problem with genetic influences?

A. The threshold required for disease expression increases as the population ages.

B. The number of affected individuals within the population has decreased.

C. The population examined for the potential problem is getting larger.

D. The threshold required for disease expression is lower.

36. Which situation is an example of modifier gene action?

A. A mother and father both have type O blood, and their son is born with type B

blood.

B. Two children in the same family both have sickle cell disease, but one child has more severe symptoms than the other.

C. Monozygotic twins separated at birth and raised in two different countries are of different heights (3 inches) and different weights (25 pounds).

D. A woman who has a gene mutation for diabetes mellitus type 2 delays the onset of the disease by exercising and maintaining a normal weight.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 4: Patterns of Inheritance Answer Section MULTIPLE CHOICE 1. ANS: B

Monogenic traits can be inherited from remote ancestors as well as more directly from our parents. Expression of monogenic traits is determined by the input of the two alleles (one from each parent) of a single gene. Because alleles segregate, meaning that normally only one allele of a pair is transmitted from each parent to any child, a single gene remains distinct and does not combine with alleles from other genes, inheritance patterns from one family generation to another can be traced.

PTS: 1 2. ANS: B

A single-gene trait is one in which the expression is determined by the input of two alleles for the gene. One allele for the single-gene trait is inherited randomly from father’s two alleles and one is inherited randomly from the mother’s two alleles for that same single-gene trait. A person with the normal number of chromosomes only inherits two alleles (one from each parent) regardless of how many possible alleles are present in the human population for this singlegene trait.

PTS: 1 3. ANS: A

Penetrance is how often a gene is expressed within a population when it is present. Some AD genes have greater penetrance than others, but it is never completely 100%. Some dominant gene alleles have “reduced” penetrance. In this case (with parentage proven by other means), one parent had to have a genotype of B/O for blood type. With reduced penetrance of the B allele, he or she did not express it in his or her phenotype. However, the gene is normal and can be transmitted to the infant who did express it in the blood type phenotype.

PTS: 1 4. ANS: A

Achondroplasia is an autosomal-dominant trait that is lethal when the genotype is homozygous.

Thus, both parents must be heterozygous A/n (with an A allele for achondroplasia and an n allele for normal stature). Normal stature (n allele) is recessive. Thus, for a child to have a normalstature phenotype from these two parents, her or her genotype would have to be n/n. As shown by the Punnett square, the chances for this couple to have a child with normal stature is 25% with each pregnancy.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 5. ANS: C

The risk for inheriting an autosomal-dominant allele from a parent who has the allele is always 50%. The inheritance risk is not changed by penetrance, but expression can be.

PTS: 1 6. ANS: A

With homozygosity of the autosomal-dominant trait, the parent only has a dominant allele to transmit. Therefore, any child this person produces will have received the trait allele and is likely to express it. Dominant traits are usually expressed and do not remain unexpressed from one generation to another. For autosomal-dominant transmission, the genders of the parent and of the child are unimportant. Both males and females transmit the allele equally to sons and daughters.

PTS: 1 7. ANS: C

Autosomal-recessive traits are only expressed in the homozygous state in which the genotype matches the phenotype. Autosomal-dominant traits can be expressed in either the homozygous or the heterozygous state. Thus, with autosomal-dominant expression, phenotype and genotype do not have to match. In X-linked-recessive traits, males express the trait in a dominant manner. In X-linked-dominant traits, females express the trait in either the homozygous or heterozygous state.

PTS: 1 8. ANS: D

Achondroplasia is an autosomal-dominant trait that is lethal when the genotype is homozygous.

Thus, both parents must be heterozygous A/n (with an A allele for achondroplasia and an n allele for normal stature). Normal stature (n allele) is recessive. Thus, for the child to have a normalstature phenotype from these two parents, her genotype is n/n. Because she does not have an A allele as her genotype, she can only transmit an n allele. The Punnett square would show:

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 9. ANS: A

Because autosomal-recessive traits or disorders are only expressed in the homozygous state, expression of the trait or disorder within a large kindred remains at a relatively constant level of no greater than about 25%. The heterozygous individual is a carrier who has a 50% chance with each pregnancy of transmitting the allele to his or her offspring. There is no evidence to support the presence of the concept of “penetrance” in autosomal-recessive traits and disorders. This issue appears to be associated only with dominant traits and disorders.

PTS: 1 10. ANS: D

Attached earlobes (versus free-swinging earlobes) is an autosomal-recessive trait. A person who expresses this trait is homozygous for the allele and transmits the allele to all his or her children regardless of the type of earlobes any of them express. This makes them obligate carriers of the allele for earlobe attachment. The hemophilia gene mutation is on the X chromosome, which a father cannot transmit to his son. Marfan syndrome and widow’s peak are autosomal-dominant traits for which there is no carrier status. If the allele is present, it is expressed.

PTS: 1 11. ANS: B

Autosomal-recessive traits and disorders must be homozygous to be expressed. Two carrier parents may be phenotypically normal although they have a heterozygous genotype because the recessive allele is not expressed in either of them. However, each can transmit the recessive allele.

Any child who inherits both recessive alleles would then express the trait or disorder. The chances for expression of the trait/disorder from parents who are heterozygous is 25% with each pregnancy.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 12. ANS: B

Recessive traits and disorders require two alleles (homozygosity) for expression. A person who has only one allele does not express the trait or disorder. Thus, the trait can “hide” for many generations but still be present in a population, although people may not be aware of its existence.

When it is finally expressed because a heterozygous carrier of the trait or disorder has children with another heterozygous carrier of the same trait or disorder, it might be considered a “random” event, especially if the couple does not have another affected child. Dominant traits are more frequently expressed because one affected allele results in expression, and they are more obvious. Therefore, if the trait or disorder is expressed in early childhood in more than one family member, people consider the possibility that it is inherited. If the trait or disorder also is considered “undesirable,” the affected individual may not be chosen as a mate for the purposes of having children. So, the key to the continued existence of recessive traits/disorders is the fact that problems are not obviously expressed in the heterozygous individual.

PTS: 1 13. ANS: C

With X-linked-dominant transmission of a trait or disorder, an affected father transmits the disorder to all of his daughters (who then express the disorder) and to none of his sons. An affected woman generally has unaffected daughters, affected daughters, unaffected sons, and affected sons in equal proportions. The most notable feature that distinguishes this transmission from that of autosomal recessive is the complete lack of father-to-son transmission because a father cannot transmit an X chromosome to his son (if he did, his son would be his daughter).

PTS: 1 14. ANS: A

It is most likely the woman’s mother (grandmother of the infant with hemophilia) who is the carrier of this X-linked abnormal gene. For autosomes, the expected percentages of maternal and paternal chromosomes inherited in any cell, organ, or tissue is 50% for each. However, because women inactivate one X chromosome randomly in all cells, which becomes “fixed” early in embryonic development, genes from the X chromosome can be unequally expressed when percentages are different for maternally derived and paternally derived X chromosomes. This woman has only 15% expression of the genes that are from paternally derived X chromosomes in the tissues that make clotting factor VIII and 85% expression of the genes that are from the maternally derived X chromosome (with the abnormal and nonfunctional gene).

PTS: 1 15. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

When an autosomal-dominant gene with variable expressivity is inherited, the gene is always expressed, but some people have more severe problems than do other people. For example, the gene mutation for one form of neurofibromatosis (NF1) is dominant. Some people with this gene mutation express it as only a few areas of light-brown skin tone known as café au lait spots. Other people, even within the same family, who have the same gene mutation develop hundreds of tumors (neurofibromas) that protrude through the skin. A person with low expression of this problem can transmit the gene to his or her child, who then may have high expression of the problem. The reverse also is true. A person with high expression can transmit the gene to his or her child, who then may have low expression of the problem. Thus, the degree of expressivity of the trait in the transmitting parent does not predict the degree of trait expressivity in the inheriting child.

PTS: 1 16. ANS: B

Choices A, C, and D have no bearing on the correct response. Regardless of whether A, C, or D are true statements or not (hair color actually is an autosomal single-gene trait), a person can only inherit traits from his or her father and mother, not from parents’ siblings. Both a person and his uncle may have inherited the same trait from a more remote common vertical ancestor (grandparent, great-grandparent) but not from horizontal ancestors (siblings, aunts, uncles, cousins).

PTS: 1 17. ANS: A

The fact that males and females are nearly equally represented within the two colors of the offspring indicates that feather color is not sex-linked (thus choices C and D are not correct) but is autosomal. If the blue allele were dominant and the yellow allele were recessive, there would be approximately three times more blue offspring (of both sexes) than yellow offspring (of both sexes). Because the yellow color is expressed in nearly a 3-to-1 ratio in the offspring, the colors follow Mendel’s rules for yellow being a dominant allele (expressed in either the homozygous state [y/y] or the heterozygous state [y/b]) and blue being a recessive allele (expressed only in the homozygous state [b/b]). Ask the students how you can determine that these two feather colors are not codominant.

Answer: the offspring would all either be green (blue and yellow make green) or a combination of blue and yellow feathered.

PTS: 1 18. ANS: C

The pattern demonstrates codominant inheritance in which both alleles are expressed equally.

The phenotype accurately expresses the genotype. Short tongues must have two short gene alleles (homozygous), medium-length tongues must have one short-tongue gene allele and one longtongue gene allele (heterozygous), and long tongues must have two long-tongue gene alleles (homozygous).

PTS: 1 19. ANS: C

Widow’s peak and taste sensitivity to PTC are inherited only in an autosomal-dominant pattern and are expressed in the homozygous and heterozygous condition. Diabetes type 2 is a complex disorder. Polycystic kidney disease has both an autosomal-dominant form and an autosomal-

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

recessive form (more severe).

PTS: 1

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20. ANS: B

Pleiotropy occurs with some single genes that have effects in more than one tissue or organ.

Usually, the pleiotropic gene codes for a substance or a structure that is found in more than one tissue or organ. Thus, a problem with that gene will result in changes that are expressed in more than one organ or body area.

PTS: 1 21. ANS: B

A highly penetrant autosomal-dominant gene is expressed close to 100% of the time and can be expressed in the homozygous or heterozygous genotype. Thus, only family members who do not express the trait or characteristic do not have the allele and cannot transmit it to their offspring.

An autosomal-dominant gene is expressed equally among males and females.

PTS: 1 22. ANS: B

Multifactorial disease results from the actions of several genes working together (polygenic) or the combined influences of genes (sometimes even single genes) and environment. Often, multifactorial traits and disorders are not dominant or recessive. Each gene variant adds to or takes away from the actual expression of the trait (phenotype). Sometimes the contributions of a particular gene variant are large, and sometimes they are small. The expression of the gene or genes is strongly affected by the environment.

PTS: 1 23. ANS: C

Huntington disease is an autosomal-dominant disorder with very high penetrance. A heterozygous person always develops the disorder, but the age of onset and the rate of disease progression can vary. This disorder does not show homozygous genotypes because homozygosity appears lethal, with loss at the embryonic or fetal pregnancy stages or within the first 12 months after birth. Thus, even if a child is born who is homozygous for the Huntington disease mutation, he or she will not live long enough to transmit the allele.

PTS: 1 24. ANS: C

Polydactyly is an autosomal-dominant single-gene trait. Thus, this baby boy did not inherit a recessive allele from each parent to express the trait. It is possible that the baby’s grandfather could be his father, but this is not the best explanation. In expressivity, the trait is always expressed, but the degree to which it is expressed can vary. This means that the baby’s father might have more than one extra digit, but if expressivity were the actual problem, the father would definitely also have an extra digit. With variation in penetrance, the gene is present but not always expressed. For polydactyly, the penetrance is about 80%. So, the father most likely does have the allele, but it is not expressed in him. However, because the allele is present, he can transmit it (with a probability of 50% for each pregnancy) to his children.

PTS: 1 25. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

When twins are monozygotic (identical), they share identical genomes at all loci. If one twin has a disease that is completely due to genetic variants, then the likelihood of the other twin having the same disease is 100%. If environment or nongenetic factors play a role, the likelihood of the second twin having the disease goes down, especially if the role of the environmental influence on the disorder is much stronger than the genetic influence.

PTS: 1 26. ANS: D

The P1 generation indicates the initial parental generation of a family or group being observed for a specific trait or traits. So, Adele and Adam are the P1 generation. The F generations are the succeeding generations of offspring or progeny produced from the parental generation.

Each succeeding generation is designated by a numeric subscript. The F1 is the first-generation offspring or progeny after the parental generation (Barbara, Brenda, and Beverly). The F2 indicates the second-generation offspring or progeny (Carol, Cliff, Carl, and Charlie). The description of this family does not include the F3 or third generation.

PTS: 1 27. ANS: A

True females have a pair of X chromosomes that determine their genetic sexual identity and do not have a Y chromosome. Therefore, females cannot inherit traits or disorders that are coded only for genes on the Y chromosome.

PTS: 1 28. ANS: D

Retinitis pigmentosa exists as genetic disorders that can be inherited in an autosomal-dominant, an autosomal-recessive, or even an X-linked-recessive mode of transmission. Syndactyly and long QT syndrome are inherited only in an autosomal-dominant transmission pattern.

Phenylketonuria is inherited only in an autosomal-recessive transmission pattern.

PTS: 1 29. ANS: C

Extremes of a condition or trait tend to become more average over time in successive generations. An IQ of 170 is very high. The average IQ is considered to be between 100 and 110.

PTS: 1 30. ANS: A

Dizygotic twins have no more common allele sequences than do any other pair of siblings, whereas monozygotic twins initially have all allele sequences in common, although some changes do occur over time. There is no overall difference in the influence of the environment over gene expression between the two twin types. Although it is possible for dizygotic twins to have different fathers, this is a very rare event and is not the main factor in the decreased concordance for expression of a monogenic trait.

PTS: 1 31. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Having a close relative (in this case a first-degree relative) who also expresses the disorder increases the likelihood that the disorder is more the result of genetic influences than environmental influences and that more risk alleles are present.

PTS: 1 32. ANS: D

The heritability estimates for these health problems are hypertension, 62%; peptic ulcer disease, 37%; congenital heart disease, 35%; schizophrenic behavior, 85%.

PTS: 1 33. ANS: A

The X chromosome is large and has about 1,500 single-gene alleles, most of which are not present on the Y chromosome or on any autosome. A few of these genes are specific for female sexual development, but there are also hundreds of gene alleles on the X chromosome that code for nonsexual functions for both males and females. Recessive alleles on the X chromosome are expressed as dominant in normal hemizygous males. The YY genotype with no X chromosome is not compatible with life. Heterozygous females are less severely affected by X-linked-dominant disorders than are males.

PTS: 1 34. ANS: C

Because the disorder is X-linked recessive, sons cannot inherit the gene or the disorder from their fathers (their fathers gave them a Y chromosome). Sons can inherit the gene from their carrier mothers, making their risk for having the disorder 50%. All daughters will at least be carriers because they must inherit the gene from their fathers. If the daughters also inherit the gene for the disorder from their mothers, they will be affected. There is no “carrier status” for sons. They are either affected or do not have the gene.

PTS: 1 35. ANS: D

A model of the liability threshold for a complex trait indicates how high the risk is in the general population and at what point risk is high enough so that having the disease or trait is likely. The top of the bell curve represents the mean of the population, or the liability (risk) of most of the population. The dotted line represents the average number of risk alleles carried by most of the population. For every trait there is a theoretical threshold, which indicates the point at which the number of risk alleles needed to express the disorder has just been met. All points to the right of the threshold line are designated as “affected.” As the line shifts to the left, more people within the study population meet the minimum threshold for expression of the complex problem.

PTS: 1 36. ANS: B

A gene’s degree of expression (disease severity) and timing of expression can be affected by modifier genes and other conditions. Although the gene abnormality that causes sickle cell disease is exactly the same for everyone with the disorder, how severely it is expressed can be changed by modifier genes (which are not always known).

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 5: Epigenetic Influences on Gene Expression Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which term or phrase is used to describe the ability of the environment to cause different

phenotypes to develop from the same genotype?

A. Developmental plasticity B. Histone modification C. Phenotype variability D. Nutrigenomics 2. How is alcohol consumption thought to increase methylation in cancer suppressor genes to

increase the risk for head and neck cancer?

A. Preventing DNA repair B. Enhancing cell division C. Acting as a methyl donor D. Activating select oncogenes 3. What would be the expected outcome for a person who has a normal gene allele and an

abnormal gene allele for insulin and the area around the abnormal gene allele is heavily methylated?

A. Normal insulin is produced in normal amounts.

B. Normal insulin is produced in lower-than-normal amounts.

C. Abnormal insulin is produced in normal amounts.

D. Abnormal insulin is produced in higher-than-normal amounts.

4. Which statement regarding the influence of epigenetic changes on gene expression is always true?

A. The change must be inherited from the parent of the same gender as the child

for expression to be affected.

B. Although gene expression is changed by inheriting these changes, the gene’s

DNA sequence remains unaffected.

C. Epigenetic changes must first occur in the parental somatic cells in order to

be inherited.

D. These changes occur throughout the genome rather than in the area of any

single specific gene.

5. What part of a chromosome’s DNA is altered by histone modification?

A. Complementary base pairing B. Attachment of phosphate groups C. Development of telomeric “caps” D. The degree of tension in helical tightness 6. What would be the expected result when the area around an oncogene is hypomethylated?

A. Cell division increases; cancer risk increases.

B. Cell division increases; cancer risk decreases.

C. Cell division decreases; cancer risk decreases.

D. Cell division decreases; cancer risk increases.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

7. By which mechanism does microRNA “silence” gene expression?

A. Preventing cytoplasm from coming into contact with the gene B. Surrounding mRNA and preventing attachment of ribosomes C. Binding to mRNA and keeping it double stranded D. Substituting a thymine for uracil 8. Which statement about the microbiome is true?

A. The microbiome does not undergo further changes after development is complete.

B. With aging, its influence on health is reduced.

C. The microbiome is necessary for good health.

D. Identical twins have identical microbiomes.

9. How is eating a diet high in broccoli thought to reduce cancer risk?

A. Increasing the excretion rate of foods known to be methyl donors B. Increasing histone acetylation, turning on anticancer genes C. Decreasing the rate of microDNA synthesis and activity D. Decreasing the exposure of carcinogens to nuclear DNA 10. How does histone modification alter gene expression?

A. Modified histones result in increased DNA methylation, which increases

the transcription of genes in that area.

B. In areas where histones are modified, the DNA is more tightly wound, and

genes are not transcribed.

C. Histone modification results in an increase in microRNA production,

which inhibits gene expression by preventing translation.

D. Histone modification results in an increase in microRNA production,

which promotes gene expression by enhancing translation.

11. In fragile X syndrome, only noncoding regions within the gene for a protein that directs and

maintains brain development are heavily methylated, resulting in decreased cognitive ability and behavioral problems. What is the most likely mechanism of this abnormal expression?

A. Increased microRNA molecules reducing the synthesis of the gene product B. Failure of the methylated regions to undergo translation C. Suppression of the “splicing out” action for introns D. Suppression of promoter function

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 5: Epigenetic Influences on Gene Expression Answer Section MULTIPLE CHOICE 1. ANS: A

Developmental plasticity is the ability of the environment to cause different phenotypes to develop from the same genotype. Histone modification is the changes of proteins around the double helix.

Phenotype variability is associated with changes in gene sequences compared with “wild-type” genes. Nutrigenomics is the study of the interaction between nutrition and the genome.

PTS: 1 2. ANS: C

Alcohol for consumption, ethyl alcohol, does have unstable methyl groups that can be donated to DNA. When these groups bind with cancer suppressor genes, their production of growthcontrolling proteins is prevented, allowing oncogene products to be unregulated.

PTS: 1 3. ANS: B

Methylation of the abnormal gene is likely to suppress its transcription, effectively blocking its production of abnormal insulin. This would allow the normal insulin allele to be expressed exclusively, which would result in the production of normal insulin. However, with only one allele active, the person would produce lower amounts of insulin.

PTS: 1 4. ANS: B

Epigenetic events affect gene expression but do not change DNA sequences. These influences are not related to the gender of the parent transmitting them. Most likely, changes occurring in the gametes (ova and sperm) or in all body cells are the events that can be transmitted to the next generation. Thus, the changes can occur throughout the genome or in areas of single genes.

PTS: 1 5. ANS: D

Histones are the proteins that DNA uses to wind around and condense to form chromosomes. In some modifications, the chromatin material keeps the DNA strands wound together so tightly, the strands cannot separate for transcription to occur. Genes in these tightly wound areas are effectively silenced and not expressed.

PTS: 1 6. ANS: A

Oncogenes produce proteins that increase cell division. If oncogenes are hypomethylated, they may not respond to suppressor gene products and be overexpressed. This action leads to excessive cell division and increases the risk for cancer development.

PTS: 1 7. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Short pieces of microRNA can bind with complementary areas of mRNA, making it double stranded. Then areas of the gene do not get “read” during translation because the ribosomes can only bind to single-stranded RNA. This results in no functional protein being synthesized.

PTS: 1 8. ANS: C

A person’s microbiome is protective in nature and is important for good health, such as helping with food digestion, keeping some pathogenic organisms in check, and stimulating immune function. The microbiome is constantly manipulated by the environment, and its influence on health does not decrease (it may, in fact, increase). Identical twins share the same genome as well as the same microbiome at birth. Most identical twins share the same environment both in utero and shortly after birth, but as they grow older, their environments differ.

PTS: 1 9. ANS: B

Broccoli contains sulforaphane. This substance increases histone acetylation that can turn on cancer suppressor genes. Increased activity of these genes can prevent oncogene overexpression and reduce cancer risk.

PTS: 1 10. ANS: B

Histone modification involves changes to the protein structures that allow organized winding of the DNA double helix to form chromosomes. Chemical tags attach to the “tails” of the histones and can alter how tightly the DNA is packaged by adjusting the tension with which it winds. When the DNA is wound tightly around the histones, some sequences of DNA may not be available for transcription, so no protein will be made from that sequence. As a result, the gene is not expressed.

PTS: 1 11. ANS: D

Noncoding regions of some genes contain promoters that signal appropriate transcription.

When promoters are silenced by methylation, the protein is not produced even though the sequence of gene-coding regions is normal. In this case, with less (if any) production of the protein that is important for brain development and maintenance, brain development is abnormal.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 6: Autosomal Inheritance and Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which syndrome or condition represents monosomy?

A. Cri du chat B. Patau syndrome C. Turner syndrome D. Robertsonian translocation 2. Which autosomal condition or syndrome commonly features a cleft lip and palate?

A. Angelman syndrome B. Cri du chat C. Edward syndrome D. Prader–Willi syndrome 3. Which description is the best example of “genomic imprinting”?

A. A child inherits a trait that his paternal grandfather expressed but that his father

did not express.

B. Boys can inherit only masculine traits from their fathers because women do not

have a Y chromosome.

C. There is a qualitative difference in some gene alleles based on whether they are inherited from the mother or the father.

D. When the number of sex chromosomes is greater than normal, the resulting individual is most often infertile.

4. Why does an embryo with a balanced translocation have the same chance for normal

development that an embryo with a normal karyotype has?

A. The translocation stimulates DNA repair mechanisms to correct the problem before commitment occurs.

B. The embryo’s somatic cells are unaffected, expressing a normal karyotype; only germline cells express the translocation.

C. Although the location of the genetic material is abnormal with a balanced translocation, there are no missing or extra alleles.

D. When a zygote with a balanced translocation reaches the embryo stage without a spontaneous abortion, the risk for abnormal development decreases.

5. How is a Robertsonian translocation different from a reciprocal translocation?

A. A Robertsonian translocation is unbalanced, and a reciprocal translocation is

balanced.

B. A reciprocal translocation is unbalanced, and a Robertsonian translocation is

balanced.

C. Reciprocal translocations occur between whole acrocentric chromosomes, and

Robertsonian translocations occur between any two nonhomologous chromosomes.

D. Robertsonian translocations occur between whole acrocentric chromosomes, and

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

reciprocal translocations occur between any two nonhomologous chromosomes.

6. Why does a person who has a deletion of the coding regions for five genes on one chromosome

15 have some manifestations or overt problems?

A. The deletions occurred in an autosome and not in a sex chromosome.

B. He or she has only one copy of the gene alleles, which results in reduced function.

C. The overall expression of the deleted areas is the same as if trisomy 15 were present.

D. The deleted areas are introns rather than exons; thus, no compensation occurred on the homologous chromosome.

7. What is the most likely explanation for a baby boy to have Tay–Sachs disease when the

child’s mother is not a carrier for the problem, but the father is a carrier?

A. The mother is not really the biological parent of this child.

B. The mother’s normal gene allele is not expressed, and the father’s Tay–Sachs allele is expressed.

C. A new mutation occurred in the father’s sperm in which the Tay–Sachs gene has become dominant.

D. The father’s Tay–Sachs gene has been translocated from its usual locus to the Y chromosome and thus is expressed in any male child.

8. Which statement about partial deletions and duplications of autosomes is true?

A. If one child in a family has the abnormality, the risk for recurrence is 50% with

each pregnancy.

B. The aberration represents failure of chromosomal segregation during

gametogenesis.

C. The underlying cause is random chromosome breakage and rearrangement.

D. The incidence increases with advancing maternal or paternal age.

9. What is the best explanation for a person whose karyotype from blood cells shows nearly all cells

to have trisomy 21 to have 10 clinical manifestations of Down syndrome and an above-average intelligence?

A. The trisomy was a result of nondisjunction of paternal gametes instead of maternal gametes.

B. The person has pseudo–Down syndrome, in which environmental conditions caused the person to have development that mimics only the physical manifestations.

C. The person has genomic imprinting, in which the paternal number 21 chromosome is not expressed, and both maternally derived number 21 chromosomes are expressed.

D. The person has mosaicism of trisomy 21, with blood cells having a high proportion and neurons having a low proportion of cells, with three number 21 chromosomes.

10. Which reproductive consequence is possible for a man who has a 13;15 Robertsonian

translocation when his wife has a normal karyotype?

A. He is unlikely to ever impregnate his wife.

B. Because his wife has a normal karyotype, there are no reproductive consequences.

C. He has an increased risk for having a child born with either trisomy 13 or trisomy

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

15.

D. He has an increased risk for having a child born with either monosomy 13 or monosomy 15.

11. Why is Down syndrome the most common trisomy disorder seen among liveborn infants?

A. The extra alleles from trisomies of larger chromosomes are embryo lethal.

B. Nondisjunction occurs more frequently for chromosome 21 than for any other

chromosome.

C. Trisomies resulting from nondisjunction express fewer alleles than trisomies that

result from unbalanced translocations.

D. The incidence may not be truly higher but is recognized more often because the

features of a baby with trisomy 21 are more obvious and not missed as often as are the more subtle features of other trisomies.

12. What is the risk for a man who has Down syndrome to produce a child with Down syndrome with

a woman who has a normal karyotype?

A. 100% B. 50% C. 5% D. 0% 13. Which chromosome is trisomic in Edward syndrome?

A. 13 B. 18 C. 22 D. X 14. For a person who has any type of chromosomal abnormality, what factor(s) has/have the greatest

impact on both physical and cognitive potential?

A. Prenatal testing and diagnosis B. The number of affected family members C. Personal family environment and social interaction D. The number of chromosomes involved with the abnormality 15. Which statement regarding partial chromosomal deletions or duplications reflects the cause of

manifestations?

A. They result in either a triple dose of some gene alleles or a single dose of some alleles.

B. They usually result in fewer or milder manifestations than do trisomies or monosomies.

C. The manifestations are unpredictable because they commonly vary depending on parental origin.

D. General prenatal chromosomal analysis has resulted in fewer children being born with these problems.

16. Which cancer type has an increased incidence among children with WAGR syndrome?

A. Acute leukemia B. Brain tumors

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

C. Colorectal cancer D. Nephroblastoma 17. What is the risk for a woman who had a retinoblastoma in early childhood as a result of a partial

deletion on one number 13 chromosome to have a child who also develops the disorder?

A. 100% B. 50% C. 25% D. 0% 18. Which feature of a 15q deletion is present in both Angelman syndrome and Prader–Willi

syndrome?

A. Light skin, eye, and hair coloring B. Continuous smiling and jerky gait C. Excessive appetite and weight gain D. Progressive tremors and seizure activity 19. Which factor has the greatest influence on the expression of a genetic disorder such as Prader– Willi syndrome or Angelman syndrome through the process of genomic imprinting?

A. Paternal age at time of conception B. Maternal age at time of conception C. The gender of the parent transmitting the deletion D. The number of bases deleted from the chromosome 20. What is the best explanation of a child having the physical phenotype of Down syndrome but

having a totally normal karyotype on chromosomal analysis of blood?

A. Mosaicism of trisomy 21 in various tissues B. Inaccurate technique for chromosomal analysis C. Uniparental disomy for somatic cells but not for germ cells D. Selective chromosome loss during meiosis II of gametogenesis

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 6: Autosomal Inheritance and Disorders Answer Section MULTIPLE CHOICE 1. ANS: C

In Turner syndrome, the person is missing one X chromosome, which is monosomy X. People with Patau syndrome have an additional number 13 chromosome (trisomy 13). With cri du chat, there is a partial deletion of material on 5p, but the affected chromosome is still present.

Robertsonian translocations are balanced, with all chromosomes present, although positions are abnormal.

PTS: 1 2. ANS: B

The manifestations of cri du chat include microcephaly, cleft lip and palate, widely spaced eyes (hypertelorism), epicanthal folds, low-set ears with few folds, a small chin, a variety of heart defects, and moderately to severely reduced cognition. None of the other autosomal conditions or syndromes listed have a cleft lip and/or palate as a consistent or predictable feature.

PTS: 1 3. ANS: C

Genomic imprinting is an epigenetic event in which a gene (or gene allele) is inactivated by means other than mutation, so the DNA sequence of the gene remains normal, but its expression is inhibited. It usually it occurs during gametogenesis, which can allow identification of whether the allele is maternally inherited or paternally inherited. The effect of an imprinted gene allele from one parent means that only the nonimprinted allele from the other parent is expressed. For the most part, when the nonimprinted gene allele is normal, its sole expression is not a problem.

Problems arise with sole expression of mutated nonimprinted gene alleles. The most common example of genomic imprinting is that the same deletion of 15q can result in either Angelman syndrome or Prader–Willi syndrome, depending on which parent contributed the chromosomal deletion. This fact indicates that there is a qualitative difference in the alleles of 15q between men and women.

PTS: 1 4. ANS: C

The most important part of the beginning process in which a zygote is formed that has half its chromosomal material from its father and half from its mother is that the zygote inherits exactly the right amount of genetic material. The actual location of the genetic material is less important for normal growth, development, and function. With a balanced translocation, the right number of alleles is present (no more and no less) in both somatic and germline cells, but they are not all located in their customary places. Because this translocation is balanced, there is no specific risk either for abnormal development or miscarriage of the fetus who is a balanced translocation carrier. Neither phenotype nor physiologic function is affected. DNA repair mechanisms correct only individual base placement, not translocations.

PTS: 1 5. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A Robertsonian translocation is a specific type of balanced translocation created by the fusion of the entire long arms (q arms) of two acrocentric chromosomes with loss of the short arms (p arms). (The short arms of acrocentric chromosomes contain no gene alleles.) Reciprocal translocations are a specific type of translocation in which segments of two nonhomologous chromosomes break and are equally exchanged. The initial exchanges of material for reciprocal translocations and fusion of chromosomes for Robertsonian translocations are both balanced.

PTS: 1 6. ANS: B

Although the chromosome is present, five gene alleles are missing. Thus, the production of proteins coded for by those alleles occurs solely as a result of the expression of those alleles on the normal chromosome. With only one allele active for these five genes, the person would produce lower amounts of the gene products.

PTS: 1 7. ANS: B

It is possible for a specific gene from one parent to be silenced and not expressed even though the gene is normal through a variety of epigenetic events. Although the gene for Tay–Sachs is recessive, and usually two Tay–Sachs genes are required for expression of the disease, if a normal gene is not expressed, the other affected allele is expressed as if it were dominant.

PTS: 1 8. ANS: D

Unlike trisomy and monosomy, most partial chromosome duplications and deletions are random events that result from chromosomal breakage and structural rearrangement, usually during gametogenesis. They are not related to parental age, and because many individuals who have the duplication or deletion are not capable of reproduction, and unless the source of the ovum or sperm involved is the unbalanced haploid gamete of a balanced translocation carrier, there is little risk for passing on the aberration.

PTS: 1 9. ANS: D

Mosaicism is a condition in which two (or more) different karyotypes are consistently present in one individual. This represents that some cells have an abnormal karyotype, and others have a normal karyotype. Mosaicism can be tissue specific, meaning that some tissues can express a mixture of normal and abnormal karyotypes, and other tissues may express all or nearly all of just one type. In this case, the most likely explanation is that the person has far fewer neurons with trisomy 21.

PTS: 1 10. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

When the diploid precursor germ cells undergo meiosis, there are three possibilities. For the sperm of a man with a 13,15 translocation, the precursor cells have one completely normal chromosome 13, one completely normal chromosome 15, and a whole chromosome 13 linked to a whole chromosome 15. This precursor sperm is balanced with two normal 13s and two normal 15s. However, when the precursor sperm undergo meiosis, several different results occur. Some sperm will have one separate chromosome 13 and one separate chromosome 15. When this sperm fertilizes a normal haploid ovum, the resulting individual has the normal number of chromosomes 13 and 15. Some sperm will have a balanced translocation of one normal chromosome 13 attached to a normal chromosome 15. Pregnancies started with these sperm result in a child who is also a balanced translocation carrier. Some sperm will contain a separate chromosome 13 and the additional one that has the 15 attached. Pregnancies started with these sperm result in trisomy 13. Some sperm will contain a separate chromosome 15 and one that is also attached to the chromosome 13.

Pregnancies started with these sperm result in trisomy 15 (which is lethal). Monosomies of either 13 or 15 are theoretically possible but are not viable beyond the embryonic period.

PTS: 1 11. ANS: A

Trisomy as a result of nondisjunction can occur with any chromosome. Chromosome 21 contains few gene alleles. Trisomy of chromosomes containing more alleles is not compatible with life and is embryo lethal. Thus, the pregnancy does not progress beyond the first trimester.

PTS: 1 12. ANS: D

Males with trisomy 21 are sterile and do not produce offspring. Females with trisomy 21 have an approximate 50% chance of producing a zygote with trisomy 21 as the result of any conception.

PTS: 1 13. ANS: B

Trisomy 18, also known as Edward syndrome, involves an extra number 18 chromosome in all or most of a person’s cells. Trisomy 13 is Patau syndrome. No syndrome of trisomy 22 has been named. Trisomy X occurs but does not produce a syndrome of identifiable features.

PTS: 1 14. ANS: C

Expression of physical, cognitive, and social development for many people who have specific chromosomal abnormalities is variable, even for one type of abnormality. The ultimate potential of any affected person for normal physical and psychosocial development or for less than fully normal development cannot be predicted. Those individuals who remain in loving and stimulating home environments tend to have greater function than those who do not.

PTS: 1 15. ANS: A

Chromosome duplications in coding regions result in having a triple dose of some gene alleles.

Deletions in coding regions result in a single dose of some alleles. When the duplicated or deleted chromosome material of a gene-coding region is present in most or all of an affected person’s tissues, it is usually accompanied by many anatomic and functional problems. These problems may be as severe or more severe than those seen in common trisomies or monosomies.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chromosomal analysis of duplications and deletions is not sufficient to determine how many copies of gene alleles are affected because a duplication or deletion large enough to be visualized on chromosomal analysis must contain at least a minimum of 200,000 bases. Duplications and deletions do not have to be this large to cause anatomic and functional problems. Partial duplications and deletions are random events resulting from chromosomal breakage and structural rearrangement, usually during gametogenesis.

PTS: 1 16. ANS: D

About 40% of children expressing WAGR syndrome at birth go on to develop a specific type of kidney cancer, Wilms’ tumor (nephroblastoma), usually before age 5 years. This syndrome is not associated with the development of any other specific cancer type.

PTS: 1 17. ANS: B

This type of cancer has both a random form and an inherited form. The inherited form follows an autosomal-dominant pattern of expression. Individuals at greatest risk for this type of retinoblastoma are missing the RB gene on at least one chromosome 13. For those with a deletion on both number 13 chromosomes, the risk is nearly 100%.

PTS: 1 18. ANS: A

Skin, hair, and eye color that are lighter than those of other family members are features of both Angelman syndrome and Prader–Willi syndrome. Continuous smiling, jerky gait, and seizure activity occur with Angelman syndrome but not Prader–Willi syndrome. Excessive appetite and weight gain are associated with Prader–Willi syndrome but not Angelman syndrome.

PTS: 1 19. ANS: C

The same deletion of 15q can result in either Angelman syndrome or Prader–Willi syndrome, depending on which parent contributed the chromosomal deletion. When no maternal material from 15q(11-13) is present, and only paternal material is expressed, Angelman syndrome results.

When no paternal material from 15q(11-13) is present, and only maternal material is expressed, Prader–Willi syndrome results. Genomic imprinting is an epigenetic event that results in a qualitative difference in expression based on parental gender. The effect of an imprinted gene allele from one parent means that only the nonimprinted allele from the other parent is expressed, which is a problem if the allele is abnormal. The age of the parent and the size of the deletion are not critical factors in the expression of this genetic disorder.

PTS: 1 20. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Mosaicism is a condition in which two (or more) different karyotypes are consistently present in one individual or in which the karyotype does not match with the phenotype. Mosaicism can be tissue specific, meaning that some tissues can express a mixture of normal and abnormal karyotypes, and other tissues may express all or nearly all of just one type. In this question, only the normal karyotype is expressed in the person’s blood. Because most cytogenetic studies are performed on blood cells or skin cells, when mosaicism is found in these tissues, the percentage of cells with abnormal karyotypes may not represent the actual proportion of cells with the abnormality in other tissues.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 7: Sex Chromosome and Mitochondrial Inheritance and Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which feature is common among people who have Klinefelter syndrome (47,XXY) or a

karyotype with 47,XXX but not among people who have Down syndrome or Edward syndrome?

A. Severely reduced cognition B. Cleft palate C. Tall stature D. Infertility 2. Which single physical feature is most often associated with a 47,XXX karyotype?

A. Infertility B. Large breasts C. Female-pattern baldness D. Height greater than siblings 3. What is the risk for a man with a 47,XYY karyotype to produce a child with a 47,XYY karyotype?

A. 100% B. 50% C. 25% D. 0% 4. Why is the incidence of an extra Y chromosome in the general population unknown?

A. The phenotype is that of a normal male.

B. Most conceptions with 47,XYY are spontaneously aborted.

C. The life expectancy for males with 47,XYY is often short.

D. The majority of males with this sex chromosome abnormality are institutionalized.

5. A male patient is tall and has some gynecomastia along with a low sperm count. During infertility

testing, he was found to have a 47,XXY karyotype. Which disorder is consistent with these findings?

A. Normal male B. Turner syndrome C. Klinefelter syndrome D. Testicular feminization syndrome 6. Which health screening activity is most important for girls and women with Turner syndrome?

A. Mammography B. Test for osteoarthritis C. Blood pressure screening D. Pulmonary function testing 7. What is the most likely mechanism for a 45,X karyotype?

A. Anticipation B. Nondisjunction C. Expansion D. Heteroplasmy

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

8. Which of the following definitions accurately represents the concept of expansion?

A. The phenotype of a genetic condition is expressed with greater severity and

at earlier ages with succeeding generations.

B. The number of trinucleotide repeat sequences within the noncoding region of a specific gene allele is increased.

C. The number of copies of a specific gene allele is amplified on one chromosome of a homologous pair.

D. A specific gene allele has at least two copies of every exon within the gene.

9. What is the function of the FMR1 gene?

A. Promoting brain development and maintaining neuronal synapses B. Ensuring the expression of all other genes present on the X chromosome C. Preventing amyloid plaque deposits and formation of neuronal tangles in the brain D. Maintaining the structural integrity of the histone proteins surrounding

the X chromosome 10. At what specific location on an X chromosome is a break most likely to occur in fragile

X syndrome?

A. The centromere B. End of the p arm C. End of the q arm D. Within the nucleosome 11. Which of the following mechanisms in fragile X syndrome (FXS) leads to expression of

the disorder?

A. Trinucleotide repeat sequences interspersed with the exons of the FMR1 gene, resulting in the production of an abnormal protein B. Trinucleotide repeat sequences causing methylation of the FMR1 gene, thus silencing its expression C. Increased production of microRNA molecules that interfere with the transcription of the FMR1 gene D. Increased production of microRNA molecules that interfere with the translation of the FMR1 gene 12. Which condition or health problem is more common in women who have an FMR1 mutation?

A. Emotional instability with inappropriate expression of anger B. Patchy areas of decreased skin pigmentation C. Progressive skeletal muscle weakness D. Menopause before age 40 years 13. What would be the expected consequence to two daughter cells after mitosis if the ability to

produce mitochondria was lost?

A. Cell commitment and differentiation would fail to occur.

B. The process of protein synthesis could not be controlled.

C. The rate and amount of ATP production would be limited.

D. Future cell division would result in an uneven number of cells.

14. Which clinical feature in a newborn baby girl suggests the possibility of Turner syndrome?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Absence of Babinski reflex B. Presence of a webbed neck C. Presence of epicanthal folds D. Absence of irises of both eyes 15. Which trait or characteristic in a 10-year-old-girl would lead her pediatrician to consider the

possibility of androgen insensitivity syndrome?

A. Her ears are large and stick out.

B. She is color-blind, and her father is not.

C. She is an inch taller than her siblings were at age 10.

D. Her nipples and breasts have not started to change and develop.

16. Which type of problem results from mutations in mitochondrial DNA (mtDNA)?

A. Reduced cellular energy B. Balanced translocations C. Excessive “crossing over” D. Single-nucleotide polymorphisms 17. What factor most strongly influences the development of the paramesonephric tissue into female

sex organs?

A. Proper positioning of the bipotential gonad B. The presence of estrogen during puberty C. The presence of two X chromosomes D. The absence of a Y chromosome 18. Which process can cause a person with an XY genotype to have a female phenotype?

A. Pregnancy resulting from artificial insemination in which the semen

donor is homosexual B. Mating of a man who has Klinefelter syndrome with a woman who has

Turner syndrome C. An epigenetic disorder in which the X chromosome inactivates the Y chromosome D. Failure of gene expression resulting in a complete absence of androgen receptors 19. On which chromosome is the androgen receptor (AR) gene located?

A. 1 B. 21 C. X D. Y 20. A man and his sister are both affected with a mitochondrial disorder. Because the sister has two

children who also have the disorder, the man is concerned about the possibility of also having children with the disorder. What is the estimated risk for each pregnancy of this man producing a child with the same mitochondrial disorder that he has if his wife does not have a mitochondrial disorder?

A. 100% B. 50% C. 25% D. 0%

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 7: Sex Chromosome and Mitochondrial Inheritance and Disorders Answer Section MULTIPLE CHOICE 1. ANS: C

In childhood, the only phenotypic manifestation of Klinefelter syndrome is that the boy usually has long legs and is taller than siblings at the same ages. Often, his final height is taller than that of either parent. The most outstanding and consistent feature is that 47,XXX women are taller than average and are taller than most family members. Although 47,XXY males are infertile, women with 47,XXX are not. Neither is associated with severely reduced cognition or cleft palate.

Both Down syndrome and Edward syndrome result in shorter-than-average stature.

PTS: 1 2. ANS: D

A karyotype of 47,XXX produces a physically normal phenotype, and the individual is fertile. The most outstanding and consistent feature is that these women are taller than average and are taller than most family members. Breast development is not affected by the extra X chromosome, and neither is hair loss.

PTS: 1 3. ANS: B

Men with a 47,XYY karyotype are fertile. Because the extra Y chromosome is likely to be present in approximately 50% of mature spermatozoa, the risk for a man with this karyotype to produce a child who is also 47,XYY is also 50%.

PTS: 1 4. ANS: A

There are no consistent and specific variations in the phenotype to indicate any type of chromosomal aberration.

PTS: 1 5. ANS: C

This patient has a phenotype and karyotype consistent with Klinefelter syndrome. This is not a normal karyotype, but the patient is male. Turner syndrome karyotype is 45,X. The karyotype for testicular feminization (an old term) is 46,XY. PTS: 1 6. ANS: C

The incidence of osteoarthritis, lung problems, and breast cancer is not increased in this population. Other health problems are common and can be significant for the person who has Turner syndrome, especially long-term hypertension and hyperlipidemia, which may be present even in early childhood.

PTS: 1 7. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Inheriting only one chromosome of a pair is monosomy, which most likely occurs as a result of nondisjunction during meiosis for gamete formation. (Recall that with nondisjunction of a chromosome pair, the pair fails to separate during meiosis. Thus, there is no segregation; both chromosomes of the pair move into one gamete, and the other gamete does not receive a chromosome from this pair.) This results in an ovum with two X chromosomes and an ovum with no X chromosome. If the ovum with no X is fertilized and results in a pregnancy, the product of the pregnancy would have Turner syndrome with a karyotype of 45,X. Nondisjunction appears to occur more often with the sex chromosome, particularly the X. PTS: 1 8. ANS: B

Expansion is an increase, usually of trinucleotide repeat sequences within a gene. The definition of response A is that of anticipation. The definition of response C is amplification.

Response D is just a mutation that would result in either an inappropriate protein or no protein produced.

PTS: 1 9. ANS: A

The FMR1 gene, which is located on the X chromosome, codes for a protein that is critical in brain development and maintenance of neural synapses. If this protein is not produced in sufficient amounts or is abnormal, reduced cognition occurs, especially in males. FMX1 stands for fragile X mental retardation 1. Mutations are not responsible for the amyloid plaques and neuronal tangles that develop in Alzheimer’s disease.

PTS: 1 10. ANS: C

The term fragile X comes from the fact that when chromosomal analysis is performed under certain laboratory conditions, the X chromosome often has a small break or gap at the end of the long arm.

PTS: 1 11. ANS: B

In FXS, the FMR1 gene-coding regions are normal, but the gene’s expression is silenced by large numbers of the trinucleotide sequence of CGG being repeated within the noncoding regions of the gene.

PTS: 1 12. ANS: D

Women with FMR1 mutations or trinucleotide repeat sequences that categorizes them as premutation carriers can develop fragile X–associated primary ovarian insufficiency syndrome (FXPOI), which is an early onset of ovarian failure causing the onset of menopause in women who are premutation carriers at ages under 40 years.

PTS: 1 13. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Mitochondria are organelles within a cell’s cytoplasm that are responsible for most of the generation of a high-energy chemical substance used to power cellular work. Thus, they are the “power plants” of cellular energy production. Without mitochondria in the newly generated cell, ATP production would be profoundly reduced.

PTS: 1 14. ANS: B

Girls with Turner syndrome are often identified at birth because of the presence of “classic” phenotypic features. These include a smaller-than-expected size at full-term pregnancy, neck webbing, pedal edema, and cardiac abnormalities. These children have a normal early neurologic response, including Babinski reflex. Although epicanthal folds often occur in Turner syndrome, they are not unique to this syndrome. Aniridia is associated with WGAR syndrome and is not found in Turner syndrome.

PTS: 1 15. ANS: B

The gene for color vision is on the X chromosome and is recessive. Thus, color-blindness is an X-linked-recessive disorder resulting from a mutation in the gene. If a male inherits this gene from his mother, he will express color-blindness. If a girl inherits this gene from her mother, she will only express color-blindness if her father is color-blind.

PTS: 1 16. ANS: A

Mitochondria produce most of the adenosine triphosphate (ATP) to use as the energy source to drive cellular actions. Mutations in mitochondrial DNA reduce the amount of ATP produced and affect the function of highly metabolic cells, tissues, and organs. Crossing over and translocations do not occur in mtDNA. Although single-nucleotide polymorphisms can occur in mtDNA, they are not the problem caused by mutated mtDNA. PTS: 1 17. ANS: D

The early embryonic tissue capable of developing into male sex organs, including the penis, scrotum, prostate, and the tubular system connecting the testis to the urinary system, is the mesonephric ductal tissue (wolffian glands). The tissue capable of developing into female sex organs is the paramesonephric ductal tissue (müllerian ducts). Both types of tissues are present in embryos that are 46,XX and those that are 46,XY. Under the influence of testosterone and antimüllerian factor, which are normally secreted in the 46,XY embryo, mesonephric ductal tissue differentiates into anatomic male sex structures, and the paramesonephric ductal tissue regresses and degenerates so that anatomic female sex structures do not develop. Without the presence of Y-chromosome-influenced testosterone and anti-müllerian factor, the mesonephric ducts regress and degenerate. The lack of anti-müllerian factor together with the genetic influence of autosomal gene products causes the development of the paramesonephric ductal tissues into complete anatomic female sex structures. Although the absence of the Y chromosome is not the only reason that female structures develop, it appears to be the strongest.

PTS: 1 18. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The situation resulting in a person with an XY genotype having a naturally occurring female phenotype is complete androgen insensitivity or androgen insensitivity syndrome (AIS). In this condition, all tissues, including the masculine tissues (mesonephric ductal tissues), are missing androgen receptors. The presence of the Y chromosome starts the organization of the bipotential gonad into a testis with the secretion of testosterone; however, the testosterone has no developmental influence on the mesonephric ductal tissues because they lack the receptors for binding and allowing the testosterone to change the gene activity of these cells. Thus, the mesonephric ducts regress, and the paramesonephric ducts undergo partial growth. At birth, the child has very clear female external genitalia, and this phenotype continues throughout life. The girl does go through her adolescent growth spurt and starts to develop hip curves and breasts but does not begin menstruation. She cannot become pregnant but is female in every other sense. The position of the bipotential gonad does not influence its function. Klinefelter men and Turner women are infertile. The X chromosome does not inactivate the Y chromosome.

PTS: 1 19. ANS: C

Even though men have more androgen receptors than do women, the AR gene is located on the X chromosome, not the Y. Women do produce small amounts of androgens in the adrenal gland, which need to bind to the AR receptors in order to produce their effects.

PTS: 1 20. ANS: D

All mitochondria at conception come from the egg because they are located in the egg’s cytoplasm. The sperm has no cytoplasmic mitochondria. Therefore, most mitochondrial disorders can only be inherited from a woman with the disorder and not from a man with the disorder.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 8: Family History and Pedigree Construction Multiple Choice Identify the choice that best completes the statement or answers the question.

1. What is the primary purpose of a pedigree?

A. To identify family members’ places within a kindred and describe their

medical history.

B. To establish which person within a kindred is responsible for introducing a

genetic mutation into the family.

C. To determine the specific risk of any one family member to develop or pass on

a genetic-based health problem.

D. To establish which person within a kindred is responsible for bringing the

family to the attention of a genetics professional.

2. Why is a pedigree considered an “unstable” product?

A. The person collecting the information may not be a genetics professional.

B. Some family members may have lied about their reproductive history.

C. The memories of older family members may be inaccurate.

D. The health of living family members continues to evolve.

3. What is the significance of a horizontal dashed line drawn between a male and a female?

A. The two siblings are both adopted.

B. The individuals have a casual sexual relationship and are not married.

C. The individuals have a third-degree (or later) blood relationship with each other.

D. Neither of the two individuals is a first- or second-degree relative of the proband.

4. What is the best way to ensure completeness and accuracy in constructing a pedigree?

A. Eliminating the contributed information from any family member who

is cognitively impaired.

B. Asking the oldest person in the kindred to provide the familial information.

C. Having several family members work together to develop the pedigree.

D. Using a template for drawing the symbols.

5. In the acronym “SCREEN” for formulating questions to ask that may indicate a heritable

trait or health problem, what does the second “E” signify?

A. The ethnicity of family members B. That the problem or trait is present in every generation of the family C. That the onset of the problem is present only among elderly family members D. That the trait or problem is equally represented by male and female family members 6. Which issue is considered a “red flag” for the need for referral to a genetics professional?

A. A family history of both breast and colon cancer B. Myocardial infarction in several parents or grandparents older than 64 years of age C. The presence of brothers and sisters who do not resemble any known

family member D. The presence of neurodevelopmental disorders in one or more members of the

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

kindred 7. When constructing a pedigree around a specific health problem, what is the minimal number

of generations needed to accurately assess the presence or absence of a genetic factor in disease development?

A. One B. Two C. Three D. Four 8. What does a double horizontal line between a circle symbol and a square symbol as the only

line directly connecting two individuals signify on a pedigree?

A. The individuals are parent and child.

B. The individuals are related by blood.

C. The individuals are brother and sister.

D. The individuals have mated without being married.

9. Lois has Alzheimer’s disease. She and her daughter Linda provided care for Lois’s mother

Leona when she had severe Alzheimer’s disease. Now Linda, age 57, is showing signs of impaired memory. Linda’s daughter Lucy is pregnant with her first child and has contacted a genetic counselor to discuss Alzheimer’s risk in her family. Who in this family is the proband?

A. Leona B. Lois C. Linda D. Lucy 10. Which activity would a general registered nurse be expected to perform as part of genomic care?

A. Calculating recurrence risk for parents who have just had a child

with nondisjunction Down syndrome B. Informing a patient that his test results are positive for a genetic disorder C. Obtaining an accurate family history and physical assessment data D. Requesting a consultation visit from a clinical geneticist 11. Which demographic information could have an important influence on susceptibility to

disorders that have a strong genetic component?

A. Ethnicity B. Education C. Profession D. Nutrition status 12. What is the significance of a pedigree symbol consisting of a square with a diagonal slash

mark through it?

A. Affected female B. Affected male C. Deceased female D. Deceased male 13. Why is a legend necessary when constructing a pedigree that includes a health history?

A. Gender-altering surgeries require either the drawing of both male and female

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

symbols for the same person or the use of a diamond instead of a circle or square.

B. When a pedigree includes more than three generations, the relationships of members in older (more remote) generations are less clear and must be explained.

C. The gender of individuals lost in the first trimester of pregnancy cannot be established.

D. Health problems do not have standard symbols.

14. How are Punnett squares and pedigrees different?

A. Pedigrees represent genotypes, and Punnett squares represent phenotypes.

B. Punnett squares represent genotypes, and pedigrees represent phenotypes.

C. Pedigrees can identify actual carriers of genetic mutations, whereas

Punnett squares only suggest carrier status.

D. Punnett squares can identify actual carriers of genetic mutations, whereas pedigrees only suggest carrier status.

15. What is the risk for disease expression among the offspring of a woman who has

an autosomal-recessive disorder when her partner is a carrier for the same disorder?

A. All sons will be unaffected; all daughters will be carriers.

B. All children have a 50% risk for being affected and a 50% risk for being a carrier.

C. Daughters have a 50% risk for being affected; all sons will either be affected or carriers.

D. Each child of either gender has a 50% risk of being a carrier, a 25% risk of having the disease, and a 25% risk of neither being a carrier nor having the disease.

16. Which observations in a pedigree indicate a probable autosomal-dominant (AD) trait transmission?

A. Affected males and females appear in every generation.

B. The pedigree shows only one affected individual.

C. The trait appears to “skip” generations.

D. Only males are affected.

17. Marjory tells a nurse that she has three aunts with cancer. The two on her father’s side were

diagnosed with breast and ovarian cancers at ages 42 and 36. She also has one aunt on her mother’s side who was diagnosed with breast cancer at age 70. Marjory asks if she should have genetic testing. What should the nurse tell her?

A. “Your family history indicates a high risk, and you should definitely have genetic testing.”

B. “Because no men in your family are affected, it is not inherited cancer, so you don’t need mammograms or any special screening practices.”

C. “Because your aunts were older when they got breast cancer, it was probably sporadic, and you should just have regular mammograms like everyone else.”

D. “Your family history may indicate an increased risk for breast cancer, and a genetic counselor could help determine whether you could benefit from genetic testing.”

18. A man with syndactyly, an autosomal-dominant condition, has one child who also has

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

syndactyly. His unaffected wife states, “Because the risk for having a child with this problem is 50% and our first child has the problem, the next child will not be affected.” What is the best response to this statement?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. “That is not quite true because the risk is 50% with each pregnancy, so with

all future children, you would have a 50% risk.”

B. “That is not quite true because you could also have the mutation but not show

the problem if there is variable expressivity.”

C. “That is correct, but if you have more than two children, half of them will

have syndactyly, and 50% will not have it.”

D. “That is correct, and if you only have two children, the second one will not have syndactyly.”

19. Which pair of relatives represents a first-degree relationship?

A. Grandfather and grandson B. Aunt and nephew C. Sister and brother D. Two cousins 20. How are twins indicated on a pedigree?

A. The two gender symbols are placed on top of each other.

B. Each of the gender symbols is surrounded by an inward-facing set of brackets.

C. The vertical lines to each twin begin at the same point on the parents’

horizontal mating line.

D. The one born first has a vertical line connected to the parents’ horizontal mating line; the twin born second has a vertical line extending down from the first twin’s gender symbol.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 8: Family History and Pedigree Construction Answer Section MULTIPLE CHOICE 1. ANS: A

A pedigree is a pictorial or graphic illustration of family members’ places within a family and their medical history. It can be used as one tool to begin to identify health issues that may have a genetic component. It is not used to determine “blame” for a genetic mutation. The person within a kindred who is responsible for bringing the family to the attention of a genetics professional is a member of the pedigree, but his or her identification is not the purpose of a pedigree.

PTS: 1 2. ANS: D

Completing a family history is an easy and affordable way to begin genetic screening; however, family history is not a stable thing. On the day you take your patient’s family history, there may be no one with cardiac problems. The next day, your patient’s 35-year-old brother or sister could have a myocardial infarction, changing your patient’s genetic risk for cardiovascular disease from low to high.

PTS: 1 3. ANS: B

A horizontal dashed line between two people in a pedigree indicates that the individuals have a casual sexual relationship with each other and are not married to each other. They each could be married to someone else. This union may or may not produce offspring.

PTS: 1 4. ANS: D

The best way for a family history to be as complete and accurate as possible is for as many family members as possible to construct it together using their memories, documents, and pictures. Even those who have some cognitive impairment may remember incidents or stories that trigger memories of events in other family members.

PTS: 1 5. ANS: A

The acronym SCREEN stands for some concern about diseases that might run in the family, especially problems with reproduction, early disease or death in family members, ethnicity (some genetic diseases are more common in people who are from certain ethnic groups), and nongenetic risk factors for disease.

PTS: 1 6. ANS: D

Neurodevelopmental disorders and extreme presentations of common conditions are worth investigating by a genetics professional. Breast cancer, colon cancer, and myocardial infarction in the family history may require a genetics referral, but usually only when they appear early. As discussed in Chapter 3, physical traits from remote parents can be passed down and expressed many generations later, enabling some siblings to not resemble each other or their parents.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 7. ANS: C

Unless a pedigree shows at least three generations, no supportable conclusions regarding transmission patterns can be made.

PTS: 1 8. ANS: B

A double line indicates consanguinity, or relationship by blood. It indicates a mating between two related family members. Brother and sister are connected by “sibship lines,” and parents and children are indicated by lines of descent. A single horizontal line between a circle and a square indicates a mating but not marital status between the two.

PTS: 1 9. ANS: D

The proband is the person within a family who brought the potential genetic issue to the attention of a health-care professional or genetics professional. Lucy, even though she does not have Alzheimer’s but wishes to discuss whether the disorder has a genetic component because three related parental individuals have expressed the problem, is the proband.

PTS: 1 10. ANS: C

Being able to organize a family history into a pedigree is an important skill for all health professionals and an expectation for registered nurses with a bachelor’s degree in nursing (BSN).

The American Association of Colleges of Nursing has included in its document Essentials of Baccalaureate Education for Professional Nursing Practice 2008 that all BSN graduates must be able to “generate a pedigree from a three-generation family history using standardized symbols and terminology.” It is important that the nurse not attempt to provide genetic counseling, as this is beyond his or her scope of practice; however, it is helpful to have a good sense of when referral should be considered even though a general registered nurse does not directly perform referrals to genetic professionals.

PTS: 1 11. ANS: A

Ethnicity is critical personal information to determine because some genetic diseases are more common in people who are from certain ethnic groups. For example, Tay–Sachs disease and Gaucher disease are at least 10 times more common among people of Ashkenazi Jewish heritage (regardless of what religion they practice) than among non-Jews. A person’s ethnicity does not change and cannot be modified. Education, profession, and nutrition status have little, if any, influence on the expression of a genetic disorder.

PTS: 1 12. ANS: D

Circles are used to represent women, and squares represent men. A diagonal slash through a gender symbol indicates that the person is deceased.

PTS: 1 13. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The standardized symbols for pedigrees include gender, relationships, birth order, adoption (in or out), and matings. There are no standard symbols for any specific type of genetic, sporadic, or environmental-related health problem. The person developing the pedigree is obligated to provide a legend so that anyone who reads the pedigree will know the family’s health history.

PTS: 1 14. ANS: B

Punnett squares are diagrams that are used to determine the risk of offspring being affected when the mode of transmission and the parents’ carrier status are known. It is based on genotype.

Pedigrees show relationships and expressed traits or disorders, the phenotype. Neither pedigrees nor Punnett squares by themselves can identify actual carrier status. Pedigrees can suggest carrier status but not prove it. Punnett squares require the genotypes for a specific trait to be known for both parents to demonstrate probabilities.

PTS: 1 15. ANS: B

With an autosomal disorder, males and females are affected equally. The mother has two mutated alleles and therefore will transmit a mutated allele to all offspring. The father has only one mutated allele and has one normal allele. Each of his children has a 50% risk for inheriting his mutated allele and a 50% risk for inheriting his normal allele. Because all of the children will have a mutated allele from the mother, all children have a 50% risk for being affected and a 50% risk for being a carrier.

PTS: 1 16. ANS: A

Finding representation in each generation and that both males and females are affected provides strong evidence that the trait is being transmitted in an AD manner. When only males are affected, the trait is more likely to be X-linked recessive, with some women being carriers.

Recessive traits tend to appear on a pedigree as “skipped generations” because carriers typically do not have the phenotype. A single affected person is more likely found with an autosomalrecessive transmission, where two unaffected carriers have mated. The trait may not appear anywhere else in the pedigree. Of course, a new mutation of an AD trait could also cause only one person to be affected, but we are looking for the most probable answer.

PTS: 1 17. ANS: D

Marjory’s aunts on her father’s side were much younger than average and premenopausal when they were affected. This would raise concern that there might be genetic risk factors in the family.

The fact that they are on her father’s side makes no difference in genetic risk, and it is unclear whether or not Marjory should have genetic testing until she speaks with a genetics professional who can take a thorough family history and accurately discuss her risk.

PTS: 1 18. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Many parents believe that the probability of transmitting an autosomal-dominant trait to their children is 50% for the total number of children, rather than for each pregnancy. Thus, the statement the mother made is incorrect. Having the first child express an autosomal-dominant health problem does not mean there is no risk for the second child also expressing it. Also, only the risk or probability is 50%. If they had three more children, it is possible that all three could have the problem, that all three could not have the problem, or that some may have it and not the other(s).

PTS: 1 19. ANS: C

First-degree relatives are those that have a direct, single-line blood relationship within a pedigree (vertical or horizontal) with no other relative between them. Grandfather and grandson have a parent in between them. Aunt and nephew have a sibling/parent in between them. Cousins have two sibling parents in between them.

PTS: 1 20. ANS: C

The vertical lines connecting each twin to the parents begin at the same point on the parents’ horizontal mating line, making an upside-down “V” shape. When the twins are monozygotic, there is a small horizontal line connecting the twins’ vertical lines.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 9: Congenital Anomalies, Basic Dysmorphology, and Genetic Assessment Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which ear location is considered “low set”?

A. The earlobes are aligned with the highest point of the upper lip.

B. The earlobes are aligned with the highest point of the bottom lip.

C. The roots of the ears are aligned with the outer canthus of the eyes.

D. The roots of the ears are aligned an inch below the outer canthus of the eyes.

2. Which term is used to describe the condition of widely spaced eyes?

A. Hypertelorism B. Hypotelorism C. Hyperoculoci D. Hypo-oculoci 3. What mechanism results in the malformation of cleft lip?

A. An abnormal developmental process B. An abnormal organization of cells C. A mechanical process D. The breakdown of an originally normal developmental process 4. How is a congenital anomaly that requires intervention or management categorized?

A. A dysmorphology B. A major anomaly C. A minor anomaly D. A disruption 5. What type of problem is Pierre-Robin, in which micrognathia begins a series of events that can

result in an obstructed airway?

A. A syndrome B. An association C. A sequence D. A dysplasia 6. Which condition in a newborn should be described as a deformation if it is the only

abnormal finding?

A. A unilateral clubfoot B. A sealed (imperforate) anus C. The absence of a thyroid gland D. The presence of six toes on each foot 7. Which anomaly constitutes clinodactyly?

A. An extra toe on one foot B. A laterally curved fifth finger C. Complete absence of fingers and toes D. Fusion of two fingers on the right hand

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

8. What is the best description of a drug prescribed to control acne that also causes severe birth

defects when it is taken during pregnancy?

A. Pleiotropic B. Teratogenic C. Carcinogenic D. Geneticomimetic 9. How does genetic assessment differ from routine physical assessment?

A. Routine physical assessment relies on personal history, and genetic

assessment relies on the recorded family history.

B. Genetic assessment requires only inspection, and routine physical assessment involves inspection, palpation, and auscultation.

C. Genetic assessment encompasses routine physical assessment and always adds the consideration of genetic causes for findings.

D. Routine physical assessment is performed by all health-care professionals, and genetic assessment is performed only by genetic professionals.

10. What is the most important action to take when assessing an infant for possible dysmorphic features?

A. Assess the phenotypes of first-degree relatives.

B. Count the number of minor anomalies present.

C. Use an atlas that includes globally diverse populations.

D. Determine whether an identified dysmorphic feature is a malformation

or a deformation.

11. You are working in a clinic, and a 4-year-old child is brought in with a history of cleft palate

repair. His parents say they want to have another child and ask you if their other children will have the same problem. They mention that Mom’s uncle Bob had a cleft lip. What do you say?

A. “There is a 3% to 5% risk that your next child will be affected.”

B. “Looking at your family history, I can tell that there is a 20% risk that your next child will be affected.”

C. “Let’s make an appointment with a genetic counselor who will help determine your next child’s risk.”

D. “You should have carrier testing to find out if your next child is at risk.”

12. Naomi and her sister have the same allele for the gene of interest; however, Naomi has cleft

lip, whereas her sister has only lip pits. What genetic process explains this difference?

A. Genomic imprinting B. Decreased penetrance C. Genetic heterogeneity D. Variable expressivity 13. When geneticists assess dysmorphology, what do they consider?

A. The “gestalt” B. The general feel and overall appearance of the patient C. Specific dysmorphic features, such as the shape of the face and the position of

the ears D. All of the above

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

14. What is the significance of the presence of frontal bossing in one parent and two children

within a family as the only physical or developmental abnormality?

A. It is an inherited normal variation of an anatomical feature.

B. All three family members should be referred to a geneticist for a formal evaluation.

C. The family is most likely descended from Neanderthal ancestors rather than from Cro-Magnon ancestors.

D. This physical anatomical malformation is a type of deformation caused by prenatal compression of soft skull areas.

15. Why should a tall teenager with Marfan syndrome avoid joining the high school basketball team?

A. The excessively long limbs have reduced bone density, increasing the risk

for fractures.

B. The disorder is associated with weakened blood vessel walls that could

rupture during strenuous activity.

C. The disorder is associated with glycogen storage deficiency, and the

individual could become seriously hypoglycemic during strenuous activity.

D. Although people with Marfan syndrome are tall, they also have poor balance

that impairs gross motor coordination, increasing the risk for injury.

16. Which statement regarding cleft lip (CL) with or without cleft palate (CP) is correct?

A. It is the rarest facial anomaly.

B. These features are not found in isolation.

C. The causes are both genetic and environmental.

D. CL without CP is a minor anomaly, and CL with CP is a major anomaly.

17. What is the best action to take when you observe the presence of a smooth philtrum on a

young child?

A. Obtain a genetics referral as soon as possible.

B. Assess for the presence of any other dysmorphic features.

C. Document the finding in the medical record as the only action.

D. Gently explain to the parents that this finding is cause for concern.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 9: Congenital Anomalies, Basic Dysmorphology, and Genetic Assessment Answer Section MULTIPLE CHOICE 1. ANS: D

The usual anatomic position for the ear is having the root of the ear located on an imaginary horizontal line extending from the outer canthus to the side of the head. Thus selection C is the “normal” anatomic position for the ears, and the position described in selection D is considered low set. Earlobe location is not used to determine ear position because it varies with ear size (length).

PTS: 1 2. ANS: A

Hypertelorism describes eyes (or any paired organ) as being too far apart or widely spaced.

Hypotelorism, which is not a medical term, would indicate too short a distance between two paired organs. Neither hyperoculoci nor hypo-oculoci is a real medical term.

PTS: 1 3. ANS: A

Cleft lip/palate is caused by the failure of the lip and/or palate tissues to fuse during development.

This mechanism is considered a malformation because it is due to a developmental problem during facial formation. The tissues making up the lip do not fuse properly, leaving a gap (or cleft).

Although a cleft lip, with or without cleft palate, is usually caused by a combination of genetic and environmental factors, it is the result of an abnormal developmental process.

PTS: 1 4. ANS: B

Major anomalies are serious and usually require medical or surgical attention because they can have life-threatening implications or have a serious cosmetic effect. A minor anomaly does not have serious functional or cosmetic consequences and may or may not be surgically corrected. A dysmorphology is usually categorized as a minor anomaly. A disruption is a process, not an anomaly.

PTS: 1 5. ANS: C

In a sequence, one anomaly starts a chain reaction that causes another problem that then causes another and then sometimes another. The anomaly that starts out Pierre-Robin is micrognathia (a small jaw, which is caused by shortening and narrowing of the mandible) in the developing fetus. This small jaw causes the tongue to be positioned backward into the throat. The posteriorly positioned tongue can cause a cleft palate and sometimes respiratory obstruction. It is a sequence of events that started with the poorly formed jaw. Although all of these features (small jaw, posteriorly displaced tongue, respiratory obstruction, and cleft palate) occur together, they were all caused by the chain of events that started with the jaw problem.

PTS: 1 6. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Although clubfeet can be part of a syndrome, a unilateral clubfoot is often associated with mechanical pressure in the uterus before birth, especially among large babies. The absence of a thyroid gland and the imperforate anus are disruptions of development. The presence of six toes (polydactyly) is a malformation.

PTS: 1 7. ANS: B

Clinodactyly is a laterally curved digit, most often the fifth digit. An extra digit is polydactyly, and fused digits are syndactyly. Absence of fingers and/or toes is adactylia.

PTS: 1 8. ANS: B

A teratogen is any agent that can cause birth defects when taken or exposed to during pregnancy.

A carcinogen is any agent that can cause cancer. Pleiotropy refers to the ability of a single gene mutation to cause problems in more than one tissue. Geneticomimetic is not an official term.

PTS: 1 9. ANS: C

All parts of a routine physical assessment are included in a genetic assessment and can be performed by general health-care professionals as well as by genetic professionals. Both routine physical assessment and genetic assessment include obtaining a personal and a family history. A genetic assessment always includes consideration of a genetic cause for findings.

PTS: 1 10. ANS: A

Sometimes what seem to be minor anomalies may not be caused by any disorder but may simply reflect an inherited group of unusual features. Although an atlas representing globally diverse populations is helpful, a first step before you say an infant or child has a dysmorphic feature or is “funny looking” is to check the appearance of other family members.

PTS: 1 11. ANS: C

As a nurse, you are expected be able to recognize basic dysmorphic features and structural anomalies that should trigger referral. However, your background is not sufficient to allow you to make risk determinations or direct what type of testing a patient or family should consider.

PTS: 1 12. ANS: D

The fact that two children in the same family have lip abnormalities strongly suggests a genetic cause. In this case, a gene mutation responsible for lip abnormalities is found and is autosomal dominant in transmission, with a reduced penetrance of 80% to 90%. (Recall from Chapter 4 that 80% to 90% penetrance means that out of 100 people who have the gene variant that causes the disease, between 10 and 20 of them will show no clinical signs.) For the mutation in this case, about 70% of people with lip pits also have cleft lip (CL) or cleft palate (CP). Some family members with the same gene variant can have only lip pits, whereas others have CL/CP, and still others have just CP. This is an example of the variable expressivity.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

13. ANS: D

An accurate dysmorphology assessment must consider every one of the listed factors.

PTS: 1 14. ANS: A

With no other physical abnormalities or developmental issues, the presence of frontal bossing in a parent and offspring represents only anatomic variation that is heritable. The issue does not require genetic counseling or investigation. It is not caused by prenatal compression or post-birth positioning. Although Neanderthal prehistoric people had very prominent foreheads and brow ridges, frontal bossing is found in people of all races, origins, and ethnicities.

PTS: 1 15. ANS: B

Marfan syndrome (MFS) is an inherited genetic connective tissue disorder in which the gene for the glycoprotein fibrillin is mutated, affecting the strength of many connective tissues. The most common manifestations include tall, lanky stature; loose or lax joints; severe myopia; and cardiovascular problems, especially mitral valve prolapse, widened aorta, and aortic aneurysm.

The cardiovascular problems can significantly shorten the individual’s life span. Dissecting aortic aneurysms and death can even occur in childhood. Strenuous exercise and heavy lifting must be avoided, as should any activity in which the chest could be hit. Thus, contact sports and those that involve running or catching a ball should be avoided.

PTS: 1 16. ANS: C

CL and/or CP can be a feature of more than 400 different conditions, making it a very common facial anomaly. The malformations can be the result of a combination of genetic and environmental factors. For example, a maternal diet low in folic acid or when the mother takes certain anticonvulsant drugs during pregnancy increases the risk for CL with or without CP. Sometimes CL/CP is found with no other anomalies. Because either CL or CP can cause significant feeding problems and also have cosmetic implications, they are both considered major anomalies.

PTS: 1 17. ANS: B

A smooth philtrum is a minor anomaly, and there is a wide range of normal variations in this structure, so the parents should not be informed that this is a cause for concern. However, it is important to continue your careful examination to see if any other dysmorphic features are present in the child. Finding multiple dysmorphic features should trigger a genetics referral; however, the presence of a smooth philtrum as the only unusual finding is not a reason for a genetics referral.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 10: Enzyme and Collagen Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Why does a newborn with a genetic enzyme deficiency have a normal phenotype at birth?

A. Metabolism during prenatal life is too slow to require full enzyme activity.

B. The deficient enzyme’s activity was performed by maternal enzymes before birth.

C. During the fetal phase of life, the newborn was not exposed to the protein that

the enzyme is responsible for degrading.

D. Although the newborn cannot synthesize the enzyme after birth, the initially stored enzyme performs its functions until the level is fully depleted.

2. Which dietary alterations are necessary to help reduce the complications of phenylketonuria (PKU)?

A. Increased intake of phenylalanine; decreased intake of tyrosine B. Increased intake of phenylalanine; increased intake of tyrosine C. Decreased intake of phenylalanine; decreased intake of tyrosine D. Decreased intake of phenylalanine; increased intake of tyrosine 3. Which ethnic group has the highest incidence of a mutation in the PAH gene?

A. Ashkenazi Jews B. French Canadians from the Quebec area C. Celts from Ireland and Scotland D. Africans from equatorial areas 4. Which manifestation is a characteristic feature of untreated phenylketonuria (PKU)?

A. Increased skin pigmentation B. Excessive urination C. Fragile bones D. Small stature 5. Which action or behavior is considered an “executive function”?

A. Hopping on one foot B. Learning the names of animals C. Deciding what to give as a present D. Counting the number of objects in a picture 6. What is the expected outcome of pregnancy for women with phenylketonuria (PKU) when the

blood levels of phenylalanine are high throughout the pregnancy?

A. Most births are post-mature.

B. There is a high incidence of infertility.

C. The infant develops PKU. D. There is a high incidence of cardiovascular birth defects.

7. Which statement regarding lysosomal storage diseases is true?

A. A gene defect makes lysosomes unable to store degraded compounds.

B. Accumulation of stored iron results in cell, tissue, and organ dysfunction.

C. Defective enzymes result in the accumulation of potentially toxic substances.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

D. Mutations in the genes coding for different types of collagen produce

substances that are toxic to brain cells.

8. If a man with Gaucher type 1 has children with a woman who is a carrier for the disorder, what is

the expected risk pattern?

A. All sons will be unaffected; all daughters will be carriers.

B. All sons will be carriers; all daughters will be affected.

C. All children will have one mutated allele and will at least be carriers; each child of either gender has a 50% risk of having the disease.

D. Each child of either gender has a 50% risk of being a carrier, a 25% risk of having the disease, and a 25% risk of neither being a carrier nor having the disease.

9. Which ethnic group has the highest incidence of Gaucher disease?

A. Ashkenazi Jews B. Asian Americans C. American Indians D. Individuals of Mediterranean descent 10. Why is it important to diagnose Gaucher type 1 disease as soon as possible after birth?

A. Enzyme therapy can reduce complications for some patients.

B. When proper dietary management is instituted early, complications can

be prevented.

C. Insulin therapy can result in prevention of the development of type 2

diabetes mellitus.

D. Prophylactic therapy with antibiotics can prevent early death from pneumonia

and other infections.

11. How does intravenous enzyme-replacement therapy for Gaucher disease reduce the organ storage

of glucosylceramide?

A. The enzyme increases the destruction of glucosamine-filled macrophages.

B. The drug acts as a cofactor, increasing the activity of the mutated enzyme.

C. The drug increases production of all blood cells in the bone marrow, including white blood cells, which have not stored glucosylceramide.

D. The enzyme is absorbed through the plasma membranes of affected cells and converts the stored glucosylceramide into its constituent molecules.

12. Which assessment finding in a 6-month-old child suggests a possible lysosomal storage disease?

A. Enlarged, palpable liver B. Weight in the 95th percentile C. Does not yet say “mama” or “dada” D. Skin tone appears slightly lighter than that of either parent 13. Which therapeutic option has been found beneficial for patients with Gaucher type 1 disease?

A. Daily ingestion of oral sapropterin hydrochloride (Kuvan) B. Intravenous enzyme replacement with alpha-L iduronidase C. Weekly phlebotomy with removal of excess red blood cells D. Intravenous enzyme replacement with imiglucerase (Cerezyme) 14. For which disorder is the drug idursulfase (Elaprase) appropriate for enzyme-replacement

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

therapy?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Hurler syndrome B. Hunter syndrome C. Gaucher type 2 D. Tay–Sachs disease 15. How are the two mucopolysaccharide disorders Hurler syndrome and Hunter syndrome different?

A. Hurler syndrome is an autosomal-dominant disorder, and Hunter

syndrome is autosomal recessive.

B. Hunter syndrome is an autosomal-dominant disorder, and Hurler

syndrome is autosomal recessive.

C. Individuals with Hurler syndrome become cognitively impaired in early

childhood, whereas those with Hunter syndrome often retain intellectual ability until later in life.

D. Individuals with Hunter syndrome become cognitively impaired in early childhood, whereas those with Hurler syndrome often retain intellectual ability until later in life.

16. Why does a person with Hurler syndrome have an enlarged abdomen?

A. The excess mucopolysaccharides accumulate inside the lysosomes within the

liver cells.

B. The excess mucopolysaccharides accumulate inside the cells of the liver and the spleen.

C. The excess glycosaminoglycans weaken the muscles of the abdomen, and all contents move forward.

D. The excess glycosaminoglycans cause the person to develop type 2 diabetes, with greatly increased abdominal fat.

17. A pair of unrelated 4-year-olds (Lester and Lucy) have mucopolysaccharidosis 1 (MPS1). Lester

has severe dysmorphic features and many skeletal anomalies. Lucy has only mildly coarse features and slight developmental delay. What is the best explanation for these differences?

A. Skewed X inactivation allowed more paternal X expression for Lucy and more maternal X expression for Lester.

B. Lucy’s mother had better prenatal care, including good diet, exercise, and vitamins, than Lester’s mother.

C. It is likely that Lucy has been misdiagnosed and really has MPS2.

D. The disorder has wide variability in expression of severity.

18. If a man with Fabry disease has children with a woman who is a carrier for the disorder, what is

the expected risk pattern?

A. All sons will be unaffected; all daughters will be carriers.

B. Sons have a 50% risk for being affected; all daughters will either be affected or carriers.

C. Daughters have a 50% risk for being affected; all sons will either be affected or carriers.

D. Each child of either gender has a 50% risk of being a carrier, a 25% risk of having the disease, and a 25% risk of neither being a carrier nor having the disease.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

19. A woman who is a carrier for Fabry disease has children with a man who does not have the

disorder. Their son has the disease, and their daughter also has some symptoms of Fabry disease even though she could only have inherited one affected allele. What is the explanation for the daughter having some symptoms of Fabry disease?

A. The girl must have a different father than her brother.

B. The daughter is seeking the same attention that is given to her brother.

C. The inactivation of one X chromosome in female cells is a totally random event.

D. In addition to inheriting one affected allele, the daughter has developed a somatic mutation.

20. What is the pathologic basis of Fabry disease?

A. Increased degradation of globotriaosylceramide B. Increased accumulation of globotriaosylceramide C. Deficiency in the number of liver lysosomes D. Excessive number of liver lysosomes 21. Which manifestation in a teenage boy whose mother is a carrier for Fabry disease suggests the

boy may have the disorder?

A. Hearing is hyperacute.

B. Opacity is present in both eyes.

C. Fasting blood glucose level is elevated.

D. Growth spurt results in a height 6 inches taller than his siblings.

22. Couples from which ethnic group would derive the greatest benefit from genetic testing for Tay–

Sachs disease?

A. Ashkenazi Jews B. Asian Americans C. French Canadians from Quebec D. Individuals of Mediterranean descent 23. Which clinical manifestation in a 6-month-old infant is most diagnostic for Tay–Sachs disease?

A. Anemia and bruising B. Enlarged liver and spleen C. Cherry red spot on the retina D. Progressive cognitive impairment 24. Which substance fails to form normally in individuals with Marfan syndrome?

A. Elastin B. Glycogen C. Collagen D. Fibrillin 25. Which health problem could be expected as a result of a gene mutation that affects the correct

production and function of type 1 collagen?

A. Failure of blood to clot after minor trauma B. Increased incidence of arterial and venous aneurysms C. Increased incidence of hearing loss among children and adults D. Restrictive lung disease from excessive stiffening of alveolar walls

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

26. A son with osteogenesis imperfecta is born to parents with no family history of the disease. What

is the most likely explanation for the son’s disorder?

A. The son is not biologically related to the mother.

B. The son is not biologically related to the father.

C. The gamete of one parent had a spontaneous mutation.

D. The son’s DNA underwent a spontaneous mutation during the second trimester of pregnancy.

27. Which manifestation is most common among individuals with osteogenesis imperfecta type I? A. Short stature B. Premature birth C. Skull deformities D. Blue-tinged sclerae 28. The mother of a teenager recently diagnosed with osteogenesis imperfecta type I (OI) asks if

the problem is related to the fact that she adhered to a vegetarian diet during pregnancy.

What is the most appropriate response?

A. Your diet is not related to this disease because it is an inherited disorder.

B. Although this problem can be inherited, low calcium levels are a major cause.

C. That is one possibility, especially because collagen requires proteins to form.

D. More likely, it is related to the fact that you were older than 35 years of age when you became pregnant.

29. What is the most common cause of death among individuals with vascular Ehlers–

Danlos (Ehlers–Danlos type IV)?

A. Respiratory impairment from kyphosis B. Skin cancer (melanoma) C. Intestinal rupture D. Liver failure Multiple Response Identify one or more choices that best complete the statement or answer the question.

30. A 12-year-old boy with Marfan syndrome complains to his nurse-practitioner, “My mother won’t

let me play football or do anything else. Isn’t there something I can do besides sitting and playing video games?” Which of the following activities may be considered less risky for this child to perform?

Select all that apply.

A. Soccer (not the goalie position) B. Riding his bike with his friends C. Playing table tennis D. Competitive swimming E. Archery F. Recreational ice-skating G. Hiking low- to moderate-difficulty trails H. Basketball

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 10: Enzyme and Collagen Disorders Answer Section MULTIPLE CHOICE 1. ANS: B

The newborn with a genetic enzyme deficiency does not have symptoms of the deficiency at birth.

The reason that symptoms often are not apparent in the newborn is that the maternal enzymes cross the placenta and perform their specific functions in the cells of the fetus. When the child is born, its access to effective maternal enzymes stops, and the enzyme deficiency begins to affect the child’s metabolism.

PTS: 1 2. ANS: D

The phenylalanine hydroxylase (PAH) is responsible for converting the amino acid phenylalanine into tyrosine. The result of a deficiency of PAH is an excess of phenylalanine and a deficiency of tyrosine. To prevent complications, phenylalanine in the diet must be reduced and carefully controlled. Because very little tyrosine is made in the body as a result of PKU, more is needed in the diet.

PTS: 1 3. ANS: C

A mutation in the PAH gene is responsible for the metabolic disease phenylketonuria (PKU).

Although this disease is more common among people whose ancestors came from Northern Europe, it is especially common among Celts from Ireland and Scotland. PKU is most rare among those whose ancestors came from Africa. There is no particular increase in PKU incidence among either Ashkenazi Jews or French Canadians.

PTS: 1 4. ANS: D

When the level of phenylalanine is not managed, brain dysfunction results in severe cognitive deficiencies and diminished motor skills. Growth retardation is present, and the skin, eyes, and hair color are lighter than those of parents or unaffected siblings, rather than increased pigmentation. The urine contains large amounts of phenylalanine, but urine volume is not increased above normal.

Bone density is not affected by the disorder.

PTS: 1 5. ANS: C

Executive functions are those behavioral functions associated with prefrontal-lobe brain activity and include problem solving, impulse control, planning, and goal-directed actions. Learning names, hopping, and counting are not related to problem solving, impulse control, planning, or goal-directed activities.

PTS: 1 6. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

With increasing life spans and cognition among people with PKU, a newer issue is the problem of an increased incidence of a variety of health problems and birth defects among infants born to PKU mothers. The infants do not have PKU; however, abnormal blood levels of amino acids, especially during embryonic life, result in a wide variety of birth defects, most commonly of the cardiovascular system. The best pregnancy outcomes for PKU mothers are achieved when phenylalanine levels are well controlled before, as well as during, pregnancy.

PTS: 1 7. ANS: C

The pathologic problem in lysosomal storage diseases is that either an enzyme is not present in the correct amounts (deficiency) or the enzyme is defective and cannot perform its functions. As a result, degradation and elimination of potentially toxic substances do not occur, and these products accumulate. Neither collagen nor iron is relevant to lysosomal storage diseases.

PTS: 1 8. ANS: C

The GBA gene, which is mutated in Gaucher type 1, is located on chromosome 1q21, and the disease is transmitted in an autosomal-recessive pattern. Therefore, there is no gender difference in risk between sons and daughters.

PTS: 1 9. ANS: A

Gaucher disease is the most common of the lysosomal storage diseases and occurs most often among the Ashkenazi Jewish population (incidence approximately 1 in 450 births) compared with non-Jewish populations (incidence approximately 1 in 40,000 to 1 in 100,000 births). It is also more common among French Canadians in the Quebec area.

PTS: 1 10. ANS: A

Early diagnosis of Gaucher type 1 disease and treatment with enzyme-replacement therapy can reduce some of the severe complications of the disorder, especially hepatosplenomegaly. The enzyme-replacement therapy is expensive and must continue throughout life. Dietary management is not appropriate for the disorder. Diabetes is not a common complication of the disease, and neither is pneumonia.

PTS: 1 11. ANS: D

In Gaucher disease there is a deficiency of the GBA gene product, an enzyme that normally breaks down glucosylceramide into sugar and fat that can be recycled for other metabolic uses.

The enzyme deficiency results in the accumulation of glucosylceramide in macrophages and some other white blood cells. Enzyme replacement acts like the naturally occurring enzyme, breaking down glucosylceramide into the usual sugar and fats. Drugs for enzyme-replacement therapy are imiglucerase (Cerezyme), taliglucerase (ELELYSO), and velaglucerase (VPRIV), PTS: 1 12. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The liver contains an enormous number of lysosomes. With lysosomal storage disorders, the deficient enzyme causes products to accumulate in all lysosomes, especially those of the liver, which greatly enlarges the organ. Weight in the 95th percentile is above average but is not suggestive of lysosomal storage disease. Most infants do not say specific words by 6 months of age. Skin tone is not affected by lysosomal storage diseases, and often, the skin tone of infants is initially lighter than that of parents until the child has more exposure to sunshine.

PTS: 1 13. ANS: D

Enzyme-replacement therapy with imiglucerase (Cerezyme), which is infused intravenously every 2 weeks once a blood level has been achieved, can reduce liver size, spleen size, and bone pain within a few weeks of beginning the therapy, although each patient’s response is variable. It is one of three drugs used for enzyme-replacement therapy in Gaucher type 1. The other two drugs are taliglucerase (ELELYSO), and velaglucerase (VPRIV), PTS: 1 14. ANS: B

The enzyme defective or missing in the MPS disorder Hunter syndrome is iduronate sulfatase.

Use of the drug Elaprase can replace this enzyme. The deficient or missing enzymes in the other three disorders are not iduronate sulfatase. Thus, administering Elaprase would have no benefit.

PTS: 1 15. ANS: C

The actual pathophysiology of Hunter syndrome with regard to poor degradation of mucopolysaccharides is identical to that of Hurler syndrome. Major differences are that people with Hunter syndrome have a slower onset of symptoms, and the effect on intellectual ability is more variable. In mild forms, loss of intellectual ability is minimal. In more severe forms, the loss of intellectual ability is more severe but occurs much later than in Hurler syndrome. Hurler syndrome is an autosomal-recessive disorder resulting in deficient function of the enzyme alpha-L iduronidase, and Hunter syndrome is an X-linked recessive disorder with deficient function of the enzyme iduronate sulfatase.

PTS: 1 16. ANS: B

The excess mucopolysaccharides accumulate in all tissues and cells, accounting for the widespread changes associated with the disease. In the liver and spleen, this accumulation leads to organ enlargement, causing the abdomen to increase in size.

PTS: 1 17. ANS: D

There are over 100 different mutations in the gene that can cause Hurler syndrome (MPSI). This is responsible for the wide variations in disease severity. MPS1 is an autosomal-recessive disease, and prenatal care does not appear to have an impact on disease severity. Because the gene is not on the X chromosome, skewed X inactivation is not a plausible explanation for the difference in severity.

Clinical testing for MPS1 and MPS2 is accurate. Misdiagnosis is not likely.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

18. ANS: B

Fabry disease is an X-linked recessive disorder resulting in a deficiency of the enzyme alpha-galactosidase. Boys do not inherit an X chromosome from their fathers, only from their mothers. Therefore, an affected father cannot transmit the gene to his son. All daughters of an affected father will inherit an affected gene from him. Thus, they will all be at least carriers. If a daughter also inherits an affected X chromosome from her carrier mother, she will actually have the disorder.

PTS: 1 19. ANS: C

Although the girl may be seeking attention, she can have real symptoms of Fabry disease. In order to prevent a “double dose” of alleles on the X chromosome, one X chromosome in all somatic cells of a female is inactivated. Although Fabry disease is much more common among males, female carriers may have significant symptoms of the disorder as a result of skewed X-chromosome inactivation in different tissues, which results in greater expression of the maternal X (with the mutated allele) than the paternal X (with the normal allele). It is possible that the son’s father is not the daughter’s father, but the more likely explanation is the presence of greater inactivation of the father’s X chromosome than the mother’s X chromosome. A somatic mutation does not affect the expression of this enzyme.

PTS: 1 20. ANS: B

Fabry disease is a genetic lysosomal storage disease in which there is a deficiency of the enzyme alpha-galactosidase A (also known as ceramide trihexosidase) that results in the accumulation of globotriaosylceramide (GL-3) within the lysosomes of many tissues and organs. It does not change the number of lysosomes present in the liver (just their function).

PTS: 1 21. ANS: B

The onset of signs and symptoms usually begins later in childhood. At first, symptoms are related to poor perfusion and include cold intolerance, insufficient sweating in hot environments, and pain episodes of unknown origin. In adolescents, the symptoms worsen, with opacities developing in the eye. Deafness often occurs, not hyperacute hearing. Height is not affected or is shorter than that of unaffected siblings. Diabetes is not commonly associated with the disorder.

PTS: 1 22. ANS: A

The incidence of Tay–Sachs disease is highest among people of Ashkenazi Jewish ethnicity worldwide (1 in 3,900 births). The incidence among all other non-Jewish populations is much less frequent, about 1 in 320,000 births. French Canadians from Quebec have a higher incidence of Gaucher disease than the general population, but not of Tay–Sachs disease.

PTS: 1 23. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

During the first few months of life, the infant with Tay–Sachs disease progresses normally, learning to control the head, recognize parental faces, socially smile, and roll over. At this point, normal development slows or stops, and physical development and cognitive development regress.

The retinal cells have filled with GM2 and become pale. This makes the fovea centralis stand out as a cherry red spot against the pale retinal background. This is a hallmark of the disease.

Although cognitive ability regresses over time, this is not limited to Tay–Sachs disease. Anemia and hepatosplenomegaly are not common manifestations of the disorder.

PTS: 1 24. ANS: D

Marfan syndrome (MFS) is an inherited genetic connective tissue disorder in which the gene for the glycoprotein fibrillin is mutated. The collagen and elastin produced are normal, although the recoil strength of tissues during and after stretching is not normal because healthy fibrillin is not present to interact with the collagen and elastin. The structure and function of glycogen are not affected by the gene mutation that causes Marfan syndrome.

PTS: 1 25. ANS: B

Type 1 collagen is a major component of bones, the dermal layer of skin, tendons, ligaments, corneas, intervertebral disks, and the walls of arteries and other blood vessels. It is critical in providing strength to blood vessel walls. With insufficient or poor-quality type 1 collagen, these blood vessel walls are flabby and can balloon out and rupture.

PTS: 1 26. ANS: C

There are four major types of osteogenesis imperfecta that occur as results of mutations in a gene for type 1 collagen. All follow an autosomal-dominant transmission pattern, although spontaneous mutations are responsible about 35% of the time.

PTS: 1 27. ANS: D

The clinical manifestations of osteogenesis imperfecta type I are relatively mild and can easily be missed because the collagen produced is normal, but the amount is reduced. Usually, the person has no increase in fractures during infancy. Fractures of long bones do occur in response to relatively minor trauma throughout childhood, adolescence, and adulthood, although the skull is not often involved. In women, more fractures are seen after menopause. The most common feature is the blue-tinged coloration of the sclerae.

PTS: 1 28. ANS: A

The primary problem with OI is failure to produce at least one functional chain of procollagen that is needed to associate with other molecules and form functional collagen in bone tissue. As a result, the developing bones have less structural integrity and strength, increasing the risk for fractures. The problem is a mutation in the gene coding for type 1 collagen. It is inherited in an autosomal-dominant transmission pattern, although spontaneous mutations are responsible about 35% of the time. Diet during pregnancy does not affect the development of the disorder in a fetus who has the gene mutation.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 29. ANS: C

Vascular Ehlers–Danlos is severe and leads to premature death. The individual has very thin, fragile skin and short stature. Problems are associated with the thin connective tissue in mid-sized and large arteries, as well as in the intestinal tract. All of these tissues become thinner as the child grows. The most common causes of death are hemorrhage from arterial rupture and sepsis from intestinal rupture, often before the age of 30.

PTS: 1 MULTIPLE RESPONSE 30. ANS: B

C, E, F, G A balance of physical activity is needed to be physically fit without placing strain on the cardiovascular system. Strenuous exercise and heavy lifting must be avoided, as should any activity in which the chest could be hit. Thus, contact sports and those that involve running or catching a ball should be avoided. Walking is encouraged, as are less physically aggressive sports, such as golfing, bowling, recreational swimming, or low-intensity bicycling. Competitive swimming is too intense.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 11: Common Childhood-Onset Genetic Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which ethnic group has the highest incidence of sickle cell disease (SCD)?

A. Ashkenazi Jews B. Asian Americans C. African Americans D. French Canadians from Quebec 2. Which type of testing is most commonly used to diagnose sickle cell disease?

A. Hematocrit and hemoglobin levels B. Hemoglobin electrophoresis C. Genetic mutation analysis D. Sweat chloride analysis 3. Which specific type of genetic problem causes sickle cell disease?

A. Deletion of an exon B. Deletion of an intron C. Unbalanced translocation D. Single-nucleotide polymorphism 4. How is sickle cell trait different from sickle cell disease?

A. People with sickle cell disease are homologous for the mutation, whereas

those with sickle cell trait are heterozygous for the mutation.

B. People with sickle cell trait are homologous for the mutation, whereas those with sickle cell disease are heterozygous for the mutation.

C. Sickle cell disease results from an inherited mutation, and sickle cell trait results from an acquired mutation.

D. Sickle cell trait results from an inherited mutation, and sickle cell disease results from an acquired mutation.

5. Which feature or factor is the best predictor for delay of complications in a person who has

sickle cell disease (SCD)?

A. Male gender B. 20% or greater of HbF C. Having survived malaria D. Living in a geographic area that has cold winters 6. How does the drug Endari reduce the manifestations of sickle cell disease?

A. Promoting faster red blood cell production B. Increasing the concentration of fetal hemoglobin (HbF) C. Reducing oxidate stress and increasing the life span of red blood cells D. Correcting the mutation of one allele so that the person has sickle cell trait

instead of sickle cell disease 7. What health advantage does sickle cell trait or disease confer on the person who has it?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Decreased risk for type 1 diabetes mellitus B. Decreased risk for hypercholesterolemia C. Decreased risk for fulminating cholera D. Decreased risk for malaria 8. A woman whose hemoglobin S levels are less than 1% has a brother with sickle cell disease, and

both parents have been diagnosed as carriers for the disorder. She asks what her risks are of having a child with sickle cell disease. What is the best response?

A. “Because you do not have the trait, you cannot have a child with sickle cell disease regardless of your partner’s sickle cell status.”

B. “Because both your parents have the trait, it is possible for you to have a child with sickle cell disease if your partner actually has the disease.”

C. “Because your brother actually has sickle cell disease, the risk for your children having sickle cell disease is 50% with each pregnancy.”

D. “Because you are a woman, your daughters will each have a 50% risk for having the disease, and all of your sons will be carriers of the trait.”

9. Which ethnic group has the highest incidence of cystic fibrosis (CF)?

A. Asian Americans B. African Americans C. Hispanic Americans D. European Americans 10. Which type of testing is most commonly used to diagnose cystic fibrosis (CF)?

A. Mucus protein electrophoresis B. Genetic mutation analysis C. Sweat chloride analysis D. Stool analysis 11. Which relative of a patient who has cystic fibrosis has the correct risk for being a cystic fibrosis

carrier?

A. Sister 0% B. Mother 50% C. Father 100% D. Brother 100% 12. Which tissues are most commonly affected by mutation of the CFTR gene?

A. Eyes and ears B. Brain and intestines C. Lungs and pancreas D. Kidneys and long bones 13. What is the most likely explanation for the variability in expression of disease severity for cystic

fibrosis?

A. The ethnicity of the patient B. The specific CFTR gene mutation inherited C. The presence of other nongenetic lung or pancreatic problems D. The length of trinucleotide repeat sequences in the first exon of the CFTR gene

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

14. What health advantage does cystic fibrosis confer on the person who has it?

A. Decreased risk for type 1 diabetes mellitus B. Decreased risk for tuberculosis C. Decreased risk for anemia D. Decreased risk for typhoid 15. Which statement regarding Duchenne muscular dystrophy (DMD) is true?

A. Females are not affected.

B. Because DMD is X-linked recessive, females are affected, and males are carriers.

C. Because DMD is X-linked recessive, males are affected, and females are carriers.

D. The sons of women who are older than age 40 when pregnant are at an

increased risk for DMD. 16. Why does Duchenne muscular dystrophy (DMD) have poor reproductive fitness?

A. Most affected individuals typically do not have children.

B. Fifty percent of mutations of the dystrophin gene are deletions.

C. The loss of functional adhesion proteins prevents reproduction.

D. The expression of disease severity is highly variable among adults 17. What is the explanation for creatine kinase (CK) levels in a 14-year-old boy with Duchenne

muscular dystrophy being much lower now than they were 5 years ago?

A. His disease is improving.

B. He now performs passive rather than active exercise.

C. Most of the muscle tissue has already been destroyed.

D. The disease is probably Becker muscular dystrophy (BMD) rather than Duchenne muscular dystrophy.

18. Which additional health problem commonly develops in boys with Duchenne muscular dystrophy?

A. Arthritis B. Hypertension C. Diabetes mellitus D. Chronic heart failure 19. A 21-year-old woman who has two brothers with Duchenne muscular dystrophy asks whether

carrier status testing is appropriate for her. What is the best response?

A. “Testing could be beneficial because your risk for being a carrier is nearly 100%.”

B. “Testing could be beneficial because your risk for being a carrier is approximately 50%.

C. “Testing is not necessary because you would have expressed some symptoms of the disease by this time if you were a carrier.”

D. “Testing is not necessary because the spontaneous mutation rate for this genetic problem is high and likely to be why your brothers have the disease.”

20. Zoe, a 20-year-old woman, shows signs of mild muscle weakness. She has two first cousins on

her mother’s side who died of Duchenne’s muscular dystrophy (DMD). Which explanation regarding Zoe’s situation is most likely?

A. Zoe is a heterozygote showing partial expression B. Zoe and her cousins shared similar environmental risks.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

C. The muscle weakness is related to Zoe’s lack of exercise in her “girly-girl”

lifestyle.

D. Zoe is homozygous for DMD, with a defective dystrophin gene inherited from

both parents.

21. How is hemophilia A (classical hemophilia) different from hemophilia B? A. Symptoms of hemophilia B are not present at birth, but those of hemophilia A are.

B. In hemophilia B, the affected gene is on an autosome instead of the X

chromosome.

C. The clotting factor problem in hemophilia B is a deficiency of factor IX rather

than factor VIII. D. The physiologic effect on the partial thromboplastin time is more profound with

hemophilia than with hemophilia B. 22. Which statement regarding classic hemophilia is true?

A. Females are not affected.

B. Because hemophilia is X-linked recessive, females are affected, and males are

carriers.

C. Because hemophilia is X-linked recessive, males are affected, and females are

carriers.

D. The sons of women who are older than age 40 when pregnant are at an increased risk for hemophilia.

23. Which type of testing is most commonly used to diagnose classic hemophilia?

A. Hematocrit and hemoglobin levels B. X-chromosome inactivation C. Genetic mutation analysis D. Factor VIII levels 24. A 12-year-old girl was diagnosed with von Willebrand disease (VWD) when she developed

profound anemia from very heavy menstrual periods. Her levels of von Willebrand (vWf) factor are normal. What specific type of von Willebrand disease is she most likely to have?

A. Type 1 VWD B. Type 2 VWD C. Type 3 VWD D. Type 4 VWD 25. Which statement regarding type 1 von Willebrand disease (VWD) is true?

A. Incomplete penetrance is common.

B. It represents the female form of hemophilia.

C. Carriers can transmit the disease to their children.

D. Males are affected twice as frequently as females.

26. Which form of von Willebrand disease (VWD) is most often diagnosed in early childhood?

A. Type 1 VWD B. Type 2 VWD C. Type 3 VWD D. Type 4 VWD

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

27. Which statement about achondroplasia is true?

A. Females are affected twice as often as males.

B. The affected infant’s appearance at birth is normal.

C. The protein impaired by the mutation is the receptor for fibroblast growth factor.

D. Transmission is less of a problem among males with achondroplasia because

they are usually infertile.

28. Which health problem occurs at a higher rate among children with achondroplasia than among

the general population?

A. Cognitive impairment B. Hearing impairment C. Color blindness D. Sudden infant death syndrome 29. A son with achondroplasia is born to parents with no family history of the disease. What is the

most likely explanation for the son’s disorder?

A. The son is not biologically related to the mother.

B. The son is not biologically related to the father.

C. The gamete of one parent had a spontaneous mutation.

D. The son’s DNA underwent a spontaneous mutation during the second trimester of pregnancy.

30. Which phenotypic feature associated with achondroplasia has variable expression?

A. Disproportionally short extremities B. Larger-than-average head size C. Prominent forehead D. Hydrocephaly 31. Which factor has the greatest genetic influence in the development of type 1 diabetes mellitus?

A. Inheritance of the HLA-DR or HLA-DQ tissue types B. Sedentary lifestyle coupled with childhood obesity C. Mutation in the gene for pancreas development D. Advanced maternal age at conception 32. For individuals who have increased genetic susceptibility for type 1 diabetes, what is the most

common environmental trigger for disease expression?

A. Obesity and a sedentary lifestyle B. Exposure to radiation C. Premature birth D. Viral infection 33. Which problem or symptom is associated with type 1 diabetes but not with type 2 diabetes?

A. Insulitis B. Polydipsia C. Weight gain D. Hyperglycemia 34. Which result of genetic/genomic variation has the most support as a cause of asthma?

A. Decreased resistance to infectious microorganisms

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. Early childhood exposures to inhalation irritants (air pollution) C. Increased inflammatory responses to environmental triggers D. Mutations in frontal-lobe genes controlling attention-getting behavior

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 11: Common Childhood-Onset Genetic Disorders Answer Section MULTIPLE CHOICE 1. ANS: C

The disorder is inherited as an autosomal-recessive single-gene trait and is most common among people with African or other equatorial ancestry. The incidence of SCD in the United States among African Americans is about 1 in 400 live births. Carrier status, in which a person has only one mutated beta-globin gene allele, is estimated at 1 in 15 African Americans. Both SCD and sickle cell trait have a far greater incidence in East Africa and other equatorial countries.

PTS: 1 2. ANS: B

Genetic mutation testing is not used to diagnose SCD. It is diagnosed based on the large percentage of hemoglobin S (HbS) seen on electrophoresis. A person who has SCD usually has 80% to 90% HbS, and a person with sickle cell trait usually has less than 40% HbS. Although hemoglobin and hematocrit levels may be abnormal, they alone are not diagnostic for sickle cell disease. Sweat chloride levels are abnormal in people with cystic fibrosis and normal in people with sickle cell disease.

PTS: 1 3. ANS: D

Sickle cell disease (SCD) is a genetic disorder caused by a single-nucleotide polymorphism in both alleles of a single gene, which results in the abnormal formation of the beta chain of hemoglobin (beta globin). In the classic form of sickle cell disease, both beta-globin alleles have a mutation in which the DNA triplet coding for glutamic acid has adenine substituted for thymine (instead of CTC, the triplet now reads CAC). This mutation results in valine as the sixth amino acid in the protein sequence. There is no deletion of introns or exons, nor is there a translocation (balanced or unbalanced).

PTS: 1 4. ANS: A

When only one beta-globin allele has the mutation, the person is heterozygous for the mutation and has sickle cell trait. In this disorder, the percentage of hemoglobin S (HbS) in the red blood cells (RBCs) is usually less than 40%. When both alleles of the beta-globin gene are mutated, the person is homozygous for the mutation and has sickle cell disease. The RBCs have a high percentage, close to 90%, of HbS rather than HbA. PTS: 1 5. ANS: B

One of the best predictors for which patients who have SCD will have delayed complications is the percentage of fetal hemoglobin (HbF) that remains in circulation. HbF is a type of hemoglobin normally expressed at high levels only during fetal life, where all oxygen is derived secondhand from the maternal circulation. It tolerates low-oxygen conditions well without sickling. Higher percentages of HbF dilute the percentage of HbS and result in better tolerance of conditions that could cause sickling.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 6. ANS: C

Endari is a new drug that increases the amount of glutamine present in many cells. The increased glutamine appears to reduce the amount of oxidative stress the cells experience, which increases their functional life spans. The drug does not correct genetic mutations, nor does it promote faster red blood cell production or vascular smooth muscle relaxation. It does not affect the percentage of HbF within the red blood cells.

PTS: 1 7. ANS: D

The hemoglobin change caused by the genetic mutation of SCD does reduce the susceptibility to severe malaria on exposure to or infection with the organism. This benefit occurs in both those who are homozygous for the gene mutation and in those who are heterozygous. The disorder does not provide a benefit or reduce risk for development of type 1 diabetes, hypercholesterolemia, or cholera.

PTS: 1 8. ANS: A

Sickle cell disease is an autosomal-recessive genetic disorder. This woman does not have either sickle cell disease or sickle cell trait, as evidenced by the fact that her HbS levels are less than 1%.

Thus, regardless of her partner’s sickle cell status, she will not have a child with actual sickle cell disease. However, if her partner is a carrier, any child she has with this partner has a 50% risk of having sickle cell trait. If her partner has sickle cell disease, any child she has will be a carrier of sickle cell trait.

PTS: 1 9. ANS: D

The disorder is inherited as an autosomal-recessive single-gene trait and is most common among Caucasians of Northern and Western European heritage, although it can be found among people of all races and ethnicities. The incidence of CF in the United States among Caucasians is about 1 in 3,000 live births. Carrier status, in which a person has only one mutated gene allele, is estimated at 1 in 20 to 30 Caucasian Americans.

PTS: 1 10. ANS: C

Genetic testing is not used to diagnose the homozygous expression of CF. It is diagnosed on the basis of physical manifestations and the results of the sweat chloride test. Positive results are those indicating a high concentration of sodium chloride in the person’s sweat (60 to 200 mEq/L or mmol/L) compared with the normal value (5 to 35 mEq/L). Genetic testing by direct sequencing of the CFTR gene is useful for establishing carrier status, identifying affected children prenatally, and, to some degree, predicting disease severity but is not used as primary diagnostic testing.

PTS: 1 11. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Cystic fibrosis is inherited as an autosomal-recessive single-gene trait. Therefore, in virtually all cases, both parents are carriers. A patient’s brother and sister would each have a 50% risk for being carriers.

PTS: 1 12. ANS: C

The two main organ systems affected involve epithelial cells and are the lungs and the pancreas.

The epithelial cells in these tissues produce a thick, sticky mucus as a result of poor chloride transport that, over time, plugs up the glands in these organs, causing glandular atrophy and organ dysfunction or failure. Other organs that are affected to a far lesser degree include the liver, salivary glands, and testes.

PTS: 1 13. ANS: B

The specific mutation inherited in cystic fibrosis (CF) can make a big difference in the severity of symptoms. There are well over 1,000 different mutations, and some cause milder clinical signs and symptoms. The ethnicity of the patient is not relevant. CF is an autosomal-recessive singlegene trait, making triplet repeats not relevant. Although nongenetic liver or pancreatic problems in a person with CF could make some symptoms worse, this is not a common cause of variability in the severity of CF symptoms.

PTS: 1 14. ANS: D

The high frequency of the heterozygous carrier state for cystic fibrosis (CF) among Caucasians from Northern and Western Europe suggests that being a CF carrier might have a potential advantage.

Scientists now believe people who are heterozygous or homozygous for specific common CFTR mutations have greater resistance to typhoid and to cholera toxin when exposed to these disease-causing microorganisms.

PTS: 1 15. ANS: C

DMD is an X-linked recessive disorder, and females with the gene mutation are carriers. These women often have reduced muscle strength and cardiac problems related to random Xinactivation in these tissues even though they only have one mutated gene allele. The disorder is not affected by advanced maternal age.

PTS: 1 16. ANS: A

Most patients with DMD will die in the late teenage years or early adulthood, and they do not typically have children. There is little variability in the severity of DMD. Those with milder cases are thought to have Becker muscular dystrophy. Although deletions are possible, the dystrophin gene is so large that it is an easy target for a variety of mutations. Adhesion proteins are not involved.

PTS: 1 17. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Children with DMD have hugely elevated blood levels of the enzyme creatine kinase (CK) and the protein myoglobin during their early childhood years. Both are normally present inside intact muscle cells. When the cells are damaged or die, CK and myoglobin are released into the blood.

These levels decrease as the child ages because there is a limit to how much muscle is available to be destroyed. Therefore, lower levels in older children indicate disease progression, not disease improvement. Switching to passive exercise does not decrease muscle damage. BMD has a slower rate of muscle destruction with a more consistent level of blood CK over time.

PTS: 1 18. ANS: D

Most boys with DMD develop cardiac muscle problems because these cells also rely on dystrophin to maintain their integrity. The most common problems are dilated cardiomyopathy and chronic heart failure. There are fewer problems with smooth muscle. Those that are present do not cause vasoconstriction. Thus, hypertension is not a common problem. The muscles rather than joints are affected, and arthritis does not develop. There is no known association of DMD with the development of diabetes.

PTS: 1 19. ANS: B

The genetic mutation for Duchenne muscular dystrophy is on the X chromosome and is transmitted in an X-linked recessive manner. Although the spontaneous mutation rate for this disease is relatively high, the fact that she has two brothers with the disorder, rather than just one, strongly suggests this mutation was inherited, not spontaneous. Thus, her mother is a carrier of an X-linked recessive genetic mutation. This young woman’s risk for having inherited the mutation on her maternally derived X chromosome is 50%, not 100%. If she were a carrier, she may or may not have any symptoms of the disorder depending on which X is inactivated in her skeletal muscle and cardiac muscle tissues. The absence of symptoms does not clearly rule out a carrier status for her.

PTS: 1 20. ANS: A

Some female carriers of DMD show partial gene expression and can demonstrate symptoms of the disease. It is very likely that she is a carrier because this X-linked recessive disease has been seen in her family. It is highly unlikely that she would have received two defective dystrophin genes from her parents because men with DMD tend not to father children.

PTS: 1 21. ANS: C

Hemophilia B is a less common form of the disorder resulting from a mutation in the gene for clotting factor IX rather than for clotting factor VIII. This gene is also located on the X chromosome, and the inheritance pattern is the same as for hemophilia A. Both disorders show few if any problems right after birth because maternal clotting factors are still circulating in the newborn. Other symptoms and issues are essentially the same.

PTS: 1 22. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The F8 gene is located on the X chromosome (Xq28). Because males have only one X chromosome, which is inherited exclusively from their mothers, hemophilia has a sex-linked (Xlinked) transmission pattern. Males are either affected or unaffected; they cannot be carriers.

Few females are born with classic hemophilia because the disease is so severe in the homozygous state that these individuals die during early embryonic life. Females can be carriers. Most female carriers of classic hemophilia have reduced levels of factor VIII and have excessive bruising and bleeding. Although the spontaneous mutation rate for the gene causing this disorder is relatively high, it is not related to advanced maternal age.

PTS: 1 23. ANS: D

Genetic testing is not used to diagnose classic hemophilia. It is diagnosed on the basis of physical manifestations and low levels of factor VIII. Hematocrit and hemoglobin levels are not specifically diagnostic for hemophilia. X-chromosome inactivation does not occur in males.

Carriers can be identified by the presence of lower-than-normal levels of factor VIII, longer-thanaverage PTT, and F8 gene sequencing to determine the presence or absence of the specific mutation identified within the family.

PTS: 1 24. ANS: B

Type 2 VWD type is associated with abnormal or defective vWf, even though the amount produced is normal. Types 1 and 3 have lower circulating levels of vWf. Currently, no type 4 VWD has been identified.

PTS: 1 25. ANS: A

The VWF gene is located on an autosome, chromosome 12. Type 1 VWD is inherited in an autosomal-dominant pattern; however, because of reduced penetrance, at times, a family pedigree can give the appearance of an autosomal-recessive pattern, although there is not a true carrier status. With reduced penetrance, a person who does not manifest the disorder can transmit the affected gene to his or her children, who then may express the disorder. Males and females are affected equally.

PTS: 1 26. ANS: C

Type 3 VWD results in extremely low levels of von Willebrand factor (vWf), and bleeding is so excessive that the infant or child may first be thought to have hemophilia. For both types 1 and 2, affected individuals may not be diagnosed during childhood because symptoms are not viewed as serious, others in the family have the same problems and have learned how to manage them, or too few serious bleeding episodes have occurred in the younger years. Currently, no type 4 VWD has been identified.

PTS: 1 27. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The gene mutation causing achondroplasia is the FGFR3 gene that codes for the fibroblast growth factor receptor 3. This gene is on chromosome 4, which is an autosome. This means that the incidence is equal for males and females. The appearance at birth is very obvious, with disproportionately short arms and legs. These proportions are maintained throughout life. The fertility of males and females with achondroplasia is not affected by the disorder.

PTS: 1 28. ANS: D

The neck is very short, with an abnormal junction between the posterior head and neck. This abnormality can compress the cervical spinal cord and is believed to be a factor in the common occurrence of sleep apnea in these individuals. In addition, infants with achondroplasia have a higher incidence of sudden infant death syndrome (SIDS) than children without achondroplasia, which may be attributable to the abnormal head and neck junction or to stenosis of the opening in the skull through which the spinal cord exits (foramen magnum). Cognitive impairment, hearing impairment, and color blindness are not consistent features of achondroplasia.

PTS: 1 29. ANS: C

Achondroplasia follows an autosomal-dominant transmission pattern, although spontaneous mutations are responsible about 75% to 80% of the time. It is very common for a child with achondroplasia to be born to parents without the disorder and who have no other family members with the problem.

PTS: 1 30. ANS: D

Most of the physical features of achondroplasia, including disproportionately short extremities, large head, and prominent forehead, are present in all individuals who have the disorder. One variable feature is hydrocephaly. This occurs in some but not all children with the disorder.

PTS: 1 31. ANS: A

Susceptibility to development of type 1 diabetes mellitus (DM) as an autoimmune problem is partially determined by inheritance of the human leukocyte antigen (HLA) genes coding for the HLA-DR and HLA-DQ tissue types located on chromosome 6. However, this complex disorder also requires an environmental interaction, most often in the form of a viral infection. Sedentary lifestyle and childhood obesity are not associated with type 1 DM, just type 2. The disorder is not related to faulty pancreatic development of genetic origin. It is not related to advanced maternal age.

PTS: 1 32. ANS: D

Type 1 diabetes is an autoimmune disorder. The risk for developing the disease is increased but appears to require an additional environmental contribution within a susceptible developmental window. Most often, that environmental trigger is a viral infection.

PTS: 1 33. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Insulitis is a problem resulting from the cause of type 1 diabetes. It is infiltration of the islet cells by white blood cells, resulting in inflammation of the islet cells. This problem does not occur in type 2 diabetes because it is not an autoimmune disease. Hyperglycemia is a feature of both types and leads to polydipsia in both types. Patients with type 1 diabetes are often underweight or normal weight.

Weight gain is associated with type 2 diabetes.

PTS: 1 34. ANS: C

Common areas of identified genomic variation associated with atopic asthma include those areas that contain genes coding for proteins that are important in regulating inflammatory reactions and immune responses. One of the strongest associations currently identified with susceptibility to asthma involves 8 to 12 different single-nucleotide polymorphisms at the 1q31 locus, encoding the gene DENND1B, which produces a protein strongly active in the work of two dendritic-type cells of the immune system, the macrophage and the mast cell. Both of these cell types are involved in nonspecific inflammatory responses and in acquired (adaptive) immune responses that develop individually in susceptible people when they are exposed to an allergen.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 12: Common Adult-Onset Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Why are complex or multifactorial disorders more commonly expressed among adults than

among children?

A. As adults age, there is more time for environmental factors to influence the expression of a genetic disorder.

B. Gene expression in adults is greater than in childhood as a result of agerelated amplification of gene copy numbers.

C. Children have better DNA repair mechanisms than adults and are more likely to correct a predisposition for a complex disorder.

D. In order for a complex or multifactorial disorder to be expressed in childhood, the child must be homozygous for the genetic mutation.

2. Why is predictive testing for Huntington disease avoided for a 4-year-old child who has

one grandparent with the disorder?

A. Unless one of the parents is positive for the gene mutation, a negative finding would be noninformative.

B. The Huntington disease mutation is poorly penetrant and may never be expressed even when inherited.

C. A 4-year-old child cannot understand or participate in the genetic counseling process.

D. There are no effective primary or secondary prevention strategies for this disorder.

3. What is the best definition of the term compound heterozygous?

A. A person who has two different mutated alleles for an autosomal-

recessive disorder B. A person whose expression of an autosomal-recessive disorder is modified

by silencing of one allele C. A person who expresses a genetic disorder having different mutations on

two nonhomologous chromosomes D. A person having a complex disorder in which there is unequal influence of

genetic and environmental contributions 4. Which relatively common adult-onset disorder or health problem is the result of a mutation

in a single gene?

A. Type 2 diabetes B. Multiple sclerosis C. Hereditary hemochromatosis D. Age-related macular degeneration 5. Your patient’s family comes from Ireland. Both her parents are carriers of gene mutations

causing hereditary hemochromatosis (HFE-HHC). What is your patient’s risk of having the clinical signs of this disorder?

A. HHC is an autosomal-recessive condition; the risk is 25%.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. HHC has incomplete penetrance; we cannot accurately predict the clinical risk.

C. HHC is a complex trait; we cannot accurately predict the clinical risk.

D. HHC is an autosomal-dominant trait; the risk is 50%.

6. Why do the clinical signs and symptoms of hereditary hemochromatosis not appear until

middle adulthood?

A. The adult over 50 has had more exposures to environmental insults over time.

B. Prolonged dietary exposure to toxins produces clinical symptoms.

C. Hereditary hemochromatosis has age-related penetrance.

D. Stored iron builds up over time, causing organ damage.

7. Why are women usually older than men before symptoms of hereditary hemochromatosis manifest?

A. Women have a counterbalancing normal gene on their second X chromosome.

B. Women lose some iron with normal menstruation during childbearing years.

C. Men have a larger muscle mass and more iron-storing capability than women.

D. Men have greater expression of the gene for hemoglobin than do women.

8. If a man with hereditary hemochromatosis has children with a woman who is a carrier for

the disorder, what is the expected risk pattern?

A. All sons will be affected; all daughters will be carriers.

B. Daughters have a 50% risk for being affected; all sons will either be affected or carriers.

C. Each child of either gender has a 50% risk of being a carrier and a 50% risk of having the disease.

D. Each child of either gender has a 50% risk of being a carrier, a 25% risk of having the disease, and a 25% risk of neither being a carrier nor having the disease.

9. What do each of the genes involved in maturity-onset diabetes of the young (MODY) have

in common?

A. They are all located on the long arm of chromosome 6.

B. They each play a role in the metabolism of glucose, or insulin action and release.

C. Although they are of different sizes, they all contain the same exon and intron sequences.

D. The genes that cause MODY are expressed only in the young and suppressed during adulthood.

10. A fasting blood glucose sample indicates that your 24-year-old male patient has hyperglycemia

that was not present during a test 6 weeks ago. What is the most likely cause of this?

A. Type 1 diabetes mellitus B. Type 2 diabetes mellitus C. Maturity-onset diabetes of the young (MODY) D. Gestational diabetes mellitus 11. Which type of maturity-onset diabetes of the young (MODY) is the most common?

A. MODY-1 B. MODY-2 C. MODY-3 D. MODY-4

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

12. Which type of diabetes has a slow onset and is often diagnosed in North American only when

a complication is present?

A. Type 1 diabetes mellitus B. Type 2 diabetes mellitus C. Maturity-onset diabetes of the young (MODY) D. Gestational diabetes mellitus 13. What can be interpreted about type 2 diabetes mellitus in monozygotic twins when it affects

only one twin 70% of the time and affects both twins 30% of the time?

A. The trait is recessive.

B. Mutation repair is incompletely penetrant.

C. Nongenetic factors can influence expression.

D. Homologous genes can undergo chromatid exchanges.

14. Which condition or factor most strongly supports a genetic basis for the development of type

2 diabetes mellitus?

A. Type 2 diabetes occurring in two cousins before age 30 years B. The disease developing in a person whose parents also have type 2 diabetes C. The fact that the incidence of disease concordance is higher in dizygotic twins than in monozygotic twins D. The fact that type 2 diabetes is more prevalent in one city than it is in another city of similar size 15. Which statement best explains the “thrifty genotype” for the high incidence of type 2 diabetes?

A. The person with a thrifty genotype has an efficient metabolism that generates

more energy from less food and is more likely to survive famine.

B. The person with a thrifty genotype conserves the energy lost from the body as

heat and requires less rest to perform the same amount of work.

C. The person with a thrifty genotype stores energy rather than expends energy so

that more energy is available for survival when food is scarce.

D. The person with a thrifty genotype feels satiety with smaller amounts of food

and is less likely to become obese, thus increasing longevity.

16. Which genetic disorder is associated with an increased risk for type 2 diabetes?

A. Achondroplasia B. Down syndrome C. Huntington disease D. Hereditary hemochromatosis 17. What is the inheritance pattern for -1 antitrypsin (ATT) deficiency?

A. Autosomal dominant B. Autosomal recessive C. Autosomal codominant D. Sex-linked recessive 18. What factors increase the clinical severity of -1 antitrypsin (ATT) deficiency?

1. Whether or not the person smokes 2. Whether or not the person uses alcohol

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

3. The severity of disease in affected relatives 4. The specific alleles of the genes inherited A. 1 and 4 B. 2 and 3 C. 2 and 4 D. 4 only 19. A 31-year-old man has been diagnosed with chronic obstructive pulmonary disease (COPD) as a

result of being homozygous for the ZZ mutation of the -1 antitrypsin (AAT) gene alleles. His wife has been tested and has the MM genotype on her AAT gene alleles. The man is worried that his three children may eventually develop COPD. What is your best response?

A. “Because your wife does not have the mutation, and neither of your parents has the disease, your children will not be affected.”

B. “Because your wife is not affected and is not a carrier, your children will have normal levels of AAT, and their risk is the same as for the general population.”

C. “Because you have the mutations, and your wife does not, your son will be at an increased risk for developing COPD, but your daughter will only be a carrier.”

D. “Because your children will each have only one mutated gene allele, their risk for COPD is primarily dependent on chronic environmental exposure to inhalation irritants.”

20. Which factor is consistent with multifactorial (complex) genetic disease?

A. Expression of the trait often involves mutations in several genes.

B. Monozygotic (identical) twin concordance is 100%.

C. Genotype consistently predicts phenotype.

D. It tends to be autosomal dominant.

21. An obese patient (body mass index of 35) states, “There is no point in changing how I eat because

it is all genetics anyway.” What is the best response?

A. “Even with a large genetic component, obesity can be reduced through a sensible diet and exercise.”

B. “Let’s gather some more family history about body height, weight, and eating patterns.”

C. “You are right; you can't change your genes. There is no point in dieting.”

D. “With your genetics, your best bet would be a low-carb Atkins-type diet.”

22. Which statement regarding autoimmune diseases is true?

A. Autoimmune disorders and diseases are more common among females than

among males.

B. Autoimmune disorders and diseases are more common among males than

among females.

C. The main characteristics of an autoimmune disorder are a reduced

inflammatory response and increased susceptibility to infection.

D. The main characteristics of an autoimmune disorder are an increased

inflammatory response and decreased susceptibility to infection.

23. Which autoimmune disorder has the greatest environmental contribution?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Stickler syndrome

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. Rheumatoid arthritis C. Multiple sclerosis D. Systemic lupus erythematosus 24. Which statement is true regarding age-related macular degeneration (AMD)?

A. It is the most common cause of peripheral vision impairment among the elderly.

B. It is transmitted in an autosomal-recessive pattern of inheritance.

C. Smoking increases the risk in those who are genetically susceptible.

D. Predictive genetic testing can identify the genes that cause AMD. 25. Multiple sclerosis is more common in people from what geographic area?

A. Western Africa B. Southeast Asia C. Mediterranean region D. Northern Europe

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 12: Common Adult-Onset Disorders Answer Section MULTIPLE CHOICE 1. ANS: A

Complex or multifactorial disorders are not single-gene conditions. They are the result of polygenic input and environmental interactions. Thus, the longer a person has been exposed to the environment, the more input the environment has on the expression of a polygenic problem.

DNA repair efficiency has no relationship to age; it is related to the quality of the repair mechanisms a person inherits. Although gene amplification can occur, it is neither an agerelated phenomenon nor a cause of adult-onset complex disorders.

PTS: 1 2. ANS: D

The Huntington disease mutation has a high penetrance, which means that a person who inherits the mutation will eventually express the disease. It also means that if a person is negative for this mutation, he or she will never develop this disease. At the present time, there is no way to prevent the disorder in a person who has the mutation. Identifying the child at this age as having the mutation provides no benefit. Although it is true that a child of this age could not understand or participate in the genetic counseling, if treatment was possible, it would be worth performing the test anyway.

PTS: 1 3. ANS: A

A compound heterozygous condition can be seen in some autosomal-recessive disorders in which two mutated alleles are needed for disease expression. In disorders such as sickle cell disease, both alleles have the identical mutation, which is a homozygous condition. In disorders such as cystic fibrosis and hereditary hemochromatosis, it is possible to have two different mutated alleles contributing to disease expression. A person with this condition is heterozygous for the disease-causing alleles.

PTS: 1 4. ANS: C

Only hereditary hemochromatosis results from mutations in one gene, the HFE gene. The other disorders are complex, with more than one gene involved, and usually require a significant environmental contribution.

PTS: 1 5. ANS: B

Although the disorder is autosomal recessive, which should confer a 25% risk when both parents are carriers, it demonstrates incomplete penetrance, which is normally an issue seen with traits that have an autosomal-dominant inheritance pattern. As a result, the risk for this patient actually expressing the disorder and having clinical symptoms cannot be predicted accurately.

PTS: 1 6. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

The cause of organ damage and symptoms in hereditary hemochromatosis is the buildup of stored iron over time. This is a single-gene disorder, and environmental factors such as diet and environmental insults do not play a significant role. Hereditary hemochromatosis does not have age-related penetrance.

PTS: 1 7. ANS: B

Women with hereditary hemochromatosis do not show signs of iron overload until a later age than men, often not until well after menopause, because the menstrual cycle provides natural and regular blood iron loss every month. The gene for hereditary hemochromatosis is on an autosome, not on the X chromosome.

PTS: 1 8. ANS: C

The disease is expressed in an autosomal-recessive pattern of inheritance, and the individual must be homozygous to express the disease. Although males express the disorder at earlier ages than females, both genders are equally expressed. Because the man is homozygous for the mutation, and the woman is heterozygous, any children he fathers have a 4 out of 4 chance (100% risk) to inherit a mutated allele from him. Each child has a 2 out of 4 chance (50% risk) to inherit a mutated allele from the carrier woman. There is no chance for children of this couple to not inherit at least one mutated allele.

PTS: 1 9. ANS: B

All the genes causing MODY are involved in the metabolism of glucose, or insulin action. They are not all located on chromosome 6 and are expressed throughout life. They have different exon and intron sequences.

PTS: 1 10. ANS: C

Although both type 1 diabetes mellitus (DM) and MODY have acute onset of hyperglycemia, this patient’s age makes it more likely to be MODY rather than type 1 DM. Type 2 has a slow and insidious onset. Males cannot have gestational DM. PTS: 1 11. ANS: C

The incidences of the four types of MODY listed are as follows: MODY-1 = about 5%, MODY-2 = about 15%, MODY-3 = about 50% to 90%, and MODY-4 is very rare.

PTS: 1 12. ANS: B

Both type 1 diabetes mellitus (DM) and MODY have a rather sudden new onset. Although gestational diabetes may take only weeks to produce symptoms, it usually is of short duration and does not cause the person to manifest the usual complications of DM. Millions of people in North America do have type 2 DM, have had it for years, and are not aware of it. The two most common ways it is discovered are when complications start to appear and become bothersome and when blood tests are drawn for other reasons, and significant hyperglycemia is found.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 13. ANS: C

Type 2 diabetes mellitus is a complex or multifactorial disorder. Although there is a genetic predisposition, which both members of a set of monozygotic twins would inherit equally, environmental influences (especially obesity and sedentary lifestyle) greatly influence the development of this disease.

PTS: 1 14. ANS: B

Although type 2 diabetes mellitus (T2DM) does not strictly follow Mendelian patterns of inheritance, the evidence for genetic input is strong. About 7% of people in the general population will get T2DM at some point in their lives, but 30% to 40% of people who have one affected parent will become affected. The risk is 70% for people who have two affected parents.

Higher concordance among dizygotic twins than among monozygotic twins supports more environmental input than genetic input. Incidence between two cities is not related to genetic influences. Cousins are not first-degree relatives. The presence of the same relatively common disorder only in two cousins within a kindred does not provide strong support for a genetic influence.

PTS: 1 15. ANS: C

During historical times when populations experienced periods of plenty followed by times of famine, those people who were better at storing fat survived to pass on their genes to their offspring. People who were not metabolically “thrifty” would be less able to store fat, and they would not survive during times when there was little food.

PTS: 1 16. ANS: B

Genetic syndromes or disorders that include an increased risk of diabetes mellitus (DM) are Down syndrome, Turner syndrome, and Klinefelter syndrome. The risk for DM among the other three genetic disorders listed is not greater than that for the general population.

PTS: 1 17. ANS: C

Deficiency in ATT is inherited in an autosomal-codominant pattern in which two different versions of the ATT gene can be expressed together and contribute to the genetic trait. The most common allele, M, is associated with normal levels of -1 antitrypsin. Two different mutated alleles are known to exist, each of which confers varying degrees of disease risk. Most individuals have two copies (MM). Disease risk and severity are related to whether two mutated alleles are inherited or only one is inherited, as well as which mutated allele(s) is/are inherited.

PTS: 1 18. ANS: A

Some ATT genotypes (such as the SZ and ZZ genotypes) confer higher risk of clinical disease.

Risk for severe disease is increased by smoking, as well as by the specific gene alleles inherited.

The severity of disease in relatives and the use of alcohol have not been shown to increase risk.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

19. ANS: D

Although all of this man’s children have inherited one mutated AAT allele from him, because his wife is not a carrier, the children are heterozygous for this mutation. The production of sufficient quantities of AAT is dependent on the inheritance of a pair of normal gene alleles because the AAT gene is recessive. When one of the pair of gene alleles is missing or mutated, and the other gene allele is normal, the person makes enough AAT to prevent COPD unless there is significant exposure to cigarette smoke and other precipitating factors. When two mutated alleles are inherited, COPD develops at a fairly young age even when the person is not exposed to cigarette smoke or other irritants.

PTS: 1 20. ANS: A

Multifactorial (complex) genetic disease involves the actions of several gene variants contributing a small amount of risk and/or the combination of genetic and environmental risk factors. If monozygotic twin concordance is 100%, then the risk is entirely genetic, and the disorder has a single-gene cause. If genotype consistently predicts phenotype, the trait has a single-gene cause, and autosomal-dominant disorders also have a single gene cause.

PTS: 1 21. ANS: A

Although genetic factors can influence metabolism and food satiety, increasing energy output and decreasing energy (calorie) intake causes weight loss.

PTS: 1 22. ANS: A

Women are much more vulnerable to autoimmune diseases than men. For example, systemic lupus erythematosus (SLE) is eight times more common among women than among men. One possible explanation for this gender difference in disease expression is that the inactive X chromosome that all women carry in each cell has a potential epigenetic effect on gene expression in the genome.

Autoimmune diseases involve overaction of the immune system and inflammatory responses, not reduced activity.

PTS: 1 23. ANS: C

Multiple sclerosis is a polygenic, autoimmune-based disorder in which people with at-risk genes have greater susceptibility to disease development. It has a higher incidence among women.

Regardless of which susceptibility genes are inherited, an environmental trigger is needed to result in disease expression.

PTS: 1 24. ANS: C

AMD affects central vision and has a complex pattern of inheritance. Predictive testing is not available because the exact genetic contributions are not yet known. The most powerful environmental risk factor for AMD is cigarette smoking.

PTS: 1 25. ANS: D

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Multiple sclerosis (MS) is most common among people whose ancestors came from Northern Europe. Although people whose ancestors came from other regions can still develop MS, it is much less likely. Some of the gene allele variants thought to increase the susceptibility are not commonly found among these other ethnic groups.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 13: Cardiovascular Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which condition or factor for coronary artery disease (CAD) suggests a greater genetic

contribution than environmental contribution to its development?

A. When a person develops CAD 5 years after quitting smoking B. When CAD occurs in a person who has normal blood pressure C. When CAD is diagnosed in a person who is younger than 50 years D. When low-density lipoprotein levels can be reduced by dietary changes alone 2. The process in which gene variants interact with other gene variants to cause disease can

be described by what word/phrase?

A. Phenotype variation B. Reduced penetrance C. Epistasis D. Variable expressivity 3. What is the most accurate classification of the common forms of coronary artery disease

and hypertension?

A. Secondary disorders caused by lifestyle choices B. Sequential genetic disorders related to age, ethnicity, and gender C. Complex disorders resulting from gene–environment interactions D. Primary disorders with an autosomal-dominant pattern of inheritance 4. What is the genetic contribution to the development of the most common forms of atherosclerosis?

A. Variation in a variety of genes, with each exerting a small effect B. Disordered mitochondrial function reducing energy production C. Genetic predisposition expressed after exposure to a viral trigger D. Single-gene mutation following autosomal-dominant inheritance 5. Your 40-year-old patient is hospitalized for a myocardial infarction, but her lipid levels are

normal. She says that she would like to have a genetic test to see why she was affected at such a young age. What do you say?

A. “Heart attacks seem to be caused by a combination of many affected genes working together as well as environmental factors. There is no single gene test that will be able to identify why this happened to you.”

B. “Let’s talk with your nurse practitioner about scheduling a test for familial hypercholesterolemia.”

C. “You really shouldn’t be concerned about your genetic risk. Because you are female, it is very low.”

D. “It was just bad luck combined with the fact that you were once a smoker.”

6. Which variable in a person with coronary artery disease (CAD) increases the likelihood of a

strong genetic influence in its expression?

A. The disorder is not present in the person’s dizygotic twin.

B. The disorder is present in a woman whose father had CAD.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

C. The affected person has eaten a vegetarian diet for the past 5 years.

D. The disorder is present in about 5% of the person’s maternal family members.

7. Which total plasma cholesterol levels are most likely to be present in people who are

homozygous for an allele that causes familial hypercholesterolemia?

A. 100 mg/dL to 200 mg/dL B. 200 mg/dL to 300 mg/dL C. 400 mg/dL to 500 mg/dL D. 600 mg/dL to 1,200 mg/dL 8. Which problem associated with Tangier disease increases the risk for coronary artery disease?

A. Low levels of triglycerides B. High levels of triglycerides C. Low levels of high-density lipoproteins (HDLs) D. High levels of low-density lipoproteins (LDLs) 9. Which physical finding in a 30-year-old man suggests the possibility of familial

hypercholesterolemia?

A. Lipomas B. Xanthomas C. Osteoarthritis D. Hemangiomas 10. For patients who have familial hypercholesterolemia, what should be the focus of teaching for

blood cholesterol reduction?

A. Eliminating animal fats from the diet B. Increasing the amount of fruit in the diet C. Exercising at least 4 to 5 hours every week D. Taking the lipid-lowering drug as prescribed 11. What is true about the gene variants that cause hypertension?

A. A few genes with major contributions have been identified.

B. Genes that code for proteins in the RAAS pathways are often involved.

C. Hypertension is always secondary to another genetic disease.

D. Polymorphisms have little or no impact on the hypertensive phenotype.

12. Your patient has been diagnosed with long QT syndrome (LQTS). What do you know about

this heart rhythm problem?

A. LQTS is a congenital genetic disease that will be evident during the first 2 years of life.

B. It is relatively rare and often caused by private mutations.

C. LQTS is treated in the same way, no matter the cause.

D. Deafness always accompanies LQTS. 13. Which health problem is the most frequent cause of sudden cardiac death among young athletes?

A. Atrial fibrillation B. Hypertrophic cardiomyopathy C. Familial hypercholesterolemia type 1 D. Romano-Ward form of long QT syndrome

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

14. Why is the mitochondrial disease MELAS, which results in stroke-like episodes in addition to

encephalopathy and lactic acidosis, only transmitted by a mother and not by an affected father?

A. The mitochondria are in the cytoplasm, and virtually all the cytoplasm comes from the egg and is thus maternally inherited.

B. This is an X-linked-recessive condition, so the father can only transmit the gene to his daughters and not to his sons.

C. Mitochondria are in the cytoplasm and are not equally divided during gamete formation in meiosis.

D. MELAS reduces the percentage of Y-carrying sperm in affected males.

15. How does factor V Leiden increase the likelihood of stroke?

A. Factor V Leiden activates protein C. B. Factor V Leiden increases thrombin formation.

C. People affected with factor V Leiden have increased blood viscosity.

D. Affected people have a type of factor V that is resistant to activated protein C. 16. Which ethnic group is at higher risk for stroke caused by factor V Leiden?

A. Ashkenazi Jews B. African Americans C. Caucasians from Northern Europe D. French Canadians from the Quebec area 17. Why is factor V Leiden disorder considered a form of thrombophilia?

A. Platelet activity is impaired.

B. Blood clots form more easily.

C. Atherosclerosis development is accelerated.

D. Excessive bleeding episodes occur in response to minor trauma.

18. Why is determining the genetic contribution to the onset of stroke difficult?

A. Stroke classification and phenotype remain heterogeneous.

B. Comorbidities mask the symptoms and delay the diagnosis.

C. Environmental risk factors have equal contribution to the problem.

D. Often the person with a stroke cannot provide accurate family information.

19. Which lysosomal storage disease is associated with an increased risk for stroke?

A. Fabry disease B. Gaucher disease C. Hurler syndrome D. Tay–Sachs disease 20. What term is used to describe the gene-to-gene interaction in which the action of one gene

modifies the expression of a different gene?

A. Epistasis B. Heterogeneity C. Genomic imprinting D. Epigenetic penetrance

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

21. What do the known monogenic disorders that result in the expression of hypertension have

in common?

A. An error in the ability of vascular smooth muscle to contract B. An error in the ability of vascular smooth muscle to relax C. Excessive kidney reabsorption of sodium D. Excessive kidney excretion of potassium 22. Which statement about atrial fibrillation is correct?

A. It is the most common genetic cardiac disorder worldwide.

B. A variety of different genes can contribute to its expression.

C. It is found exclusively in adults over 60 years old.

D. When severe, it leads to long QT syndrome.

23. Which lethal cardiac arrhythmia can occur as a complication of long QT syndrome?

A. Atrial node reentry tachycardia B. Prolonged atrial fibrillation C. Mitral valve prolapse D. Torsade de pointes

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 13: Cardiovascular Disorders Answer Section MULTIPLE CHOICE 1. ANS: C

When CAD is diagnosed in a person younger than age 50 years, the genetic contribution is much greater than environmental factors compared with people who develop CAD at a later age. Even if environmental factors are strong, they require time to change healthy coronary arteries.

PTS: 1 2. ANS: C

A gene-to-gene interaction in which one gene can modify the effects of one or more other genes (in a positive way or a negative way) is called epistasis.

PTS: 1 3. ANS: C

The most common forms of coronary artery disease (CAD) and hypertension (HTN) are considered complex (multifactorial) disorders. That means that many different gene mutations work together; some increase risk, whereas others protect against disease. In addition, environment and lifestyle factors modify the severity of disease and the chance of whether or not a person will get sick at all. Some rare cardiovascular problems have a purely genetic origin.

Although both CAD and HTN are secondary to atherosclerosis, they do have both a genetic influence as well as a hereditary (lifestyle choice) influence and are not sequential disorders.

(Remember that ethnicity and gender are genetically determined.)

PTS: 1 4. ANS: A

Atherosclerosis is a common complex or multifactorial disease. The genetic contribution for the most common form is the result of problems in lots of genes working together, with each producing a small effect. The most common form is not transmitted in an autosomal-dominant fashion, is not mitochondrial, and is not the result of a viral trigger.

PTS: 1 5. ANS: A

There are many factors, both genetic and environmental, that can increase clot formation in any area of the body, including in the coronary arteries. No single-gene mutation has been found to consistently increase the risk for myocardial infarctions or the conditions that promote them.

Therefore, this complex health problem, even when it occurs in younger people, is still multifactorial.

PTS: 1 6. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

If a woman’s father had CAD, the risk of her having CAD or a myocardial infarction (MI) is 70% higher than that of a woman whose father did not have CAD. This person’s maternal family members diagnosed with the disorder at a 5% incidence represents a much lower incidence than the national average. Dizygotic twins do not have identical gene sequences at all loci. So, the absence of the problem in a dizygotic twin does not support a strong genetic influence on expression of the disorder. A genetic problem can be modified somewhat but not to a great degree by dietary changes.

PTS: 1 7. ANS: D

Total plasma cholesterol levels in people who are homozygous for an allele that causes familial hypercholesterolemia usually range between 600 and 1,200 mg/dL. Those who are heterozygous often have levels between 300 and 400 mg/dL. The range of 100 mg/dL to 200 mg/dL is considered normal or desirable. The range of 200 mg/dL to 300 mg/dL is higher than desired but much lower than for a person who has a mutation in one or both alleles of a gene associated with familial hypercholesterolemia.

PTS: 1 8. ANS: C

High-density lipoproteins are the “good” cholesterol. Low levels of HDLs are associated with an increased risk for CAD regardless of LDL level. Tangier disease is an autosomal-codominant disorder that results in very low or absent levels of HDLs. It is not related to levels of LDL or triglycerides.

PTS: 1 9. ANS: B

Xanthomas are cholesterol deposits in the skin that form when blood cholesterol levels are very high. They are associated with familial hypercholesterolemia and can occur at any age. Common sites include the elbows and knees. Lipomas are benign fatty tumors and are not related to cholesterol levels. Hemangiomas are benign blood vessel tumors and are not related to cholesterol levels.

Although osteoarthritis more commonly occurs in older adults, it can occur in younger adults who have a traumatic joint injury. Its presence is not associated with cholesterol levels.

PTS: 1 10. ANS: D

Dietary management has been the recommended initial treatment for people with familial hypercholesterolemia, but dietary modification alone can only reduce LDL levels by about 10%, so medications are often begun early, especially the statins. The best responses are achieved when the drugs are taken exactly as prescribed. Although exercise can help lower cholesterol levels, this response requires much more than 4 to 5 hours per week.

PTS: 1 11. ANS: B

The renin–angiotensin–aldosterone system (RAAS) is a pathway that regulates blood volume by controlling water and sodium retention. Blood volume is directly related to blood pressure, meaning that blood pressure increases when blood volume increases. The RAAS pathway is complex, involving many proteins and receptors, all of which are coded for by different genes.

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Problems in any one of them can lead to primary hypertension.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 12. ANS: B

Quite a large number of gene mutations have been identified as being involved in the expression of LQTS. However, it is still a relatively rare cardiac syndrome. Some of the mutations involved are only present within one kindred, known as a private mutation, and testing of the mutation beyond the affected family is not beneficial.

PTS: 1 13. ANS: B

More than one-third of young athletes, including teenagers, who die from sudden cardiac death before age 30 have hypertrophic cardiomyopathy. Many have had some symptoms of difficulty breathing with heavy exercise but have not had cardiac testing. Atrial fibrillation occurs with aging rather than in younger adults and is not induced by exercise. Familial hypercholesterolemia can result in early-age heart attacks, but most people with the problem are very exercise intolerant and do not engage in athletic activities. Romano-Ward is a congenital cardiac problem usually diagnosed in early childhood as a tachyarrhythmia. It is also fairly rare.

PTS: 1 14. ANS: A

Mitochondria are in the cytoplasm only. A sperm does not contain cytoplasm. Thus, when an egg is fertilized, all the mitochondria in it are maternally derived, regardless of whether the zygote is a male or a female.

PTS: 1 15. ANS: D

Forming blood clots only in areas where they are needed but not in other areas requires a balance between pro-clotting forces and anti-clotting forces. One anti-clotting force is activated protein C. This substance dissolves old clots and prevents clots from forming where they are not needed by degrading clotting factor V. Factor V Leiden is a mutated form of factor V that is not easily degraded by activated protein C. As a result, clots that are not needed can form more easily.

PTS: 1 16. ANS: C

Although strokes are more common among African Americans, factor V Leiden is more common in people whose ancestors came from Northern Europe. Between 3% and 8% of people with European ancestry are heterozygous (they inherited a mutation from one parent) for factor V Leiden. About 1 in 5,000 people are homozygous (they have inherited a mutant copy from each parent).

PTS: 1 17. ANS: B

Proteins of the coagulation cascade are supposed to be degraded by activated protein C (APC) so that they do not become too large or stay around too long. In people with factor V Leiden disorder, F5 (the factor V Leiden protein) is abnormal and cannot be degraded. Thus, the person experiences a state of “hypercoagulation.” It is not related to platelet aggregation and results in blood clots rather than bleeding. It is a single-gene problem unrelated to atherosclerosis.

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PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

18. ANS: A

Strokes are heterogeneous, which makes it difficult to classify them and sort out causes.

Whenever there is a lot of variation in the phenotype, it is likely that there are different genes involved in the pathogenesis. Environmental risk factors, such as smoking, are important as well.

Comorbidities also make finding genetic causes difficult. It is most likely that the genetic contribution to stroke is the result of many genes working together, with each exerting a small effect.

PTS: 1 19. ANS: A

Fabry disease is a lysosomal storage disease that is transmitted in an X-linked-recessive manner.

It is caused by the inability to make the enzyme alpha-galactosidase A, which is needed so that people can metabolize lipids. Without this enzyme, fat metabolism is disordered, and fats accumulate, which increases the risk for strokes. Tay–Sachs disease and Hurler syndrome both result in childhood deaths from respiratory problems, and no association between either of them and stroke risk has been found. Gaucher disease is not associated with an increased risk for stroke.

PTS: 1 20. ANS: A

Epistasis is the term used to describe gene-to-gene interaction, where the action of one gene modifies the expression of another. Genetic heterogeneity means that there are different genetic causes of a trait. Genomic imprinting is an epigenetic process of silencing the expression of a gene. Epigenetic penetrance is not an official genetic term.

PTS: 1 21. ANS: C

Most of the monogenic disorders that result in the expression of hypertension affect the way the kidneys control salt, causing it to be reabsorbed and retained instead of excreted. When salt is retained, excess fluid is not excreted, which leads to hypertension. The genes involved are usually part of complex pathways, such as the renin–angiotensin–aldosterone (RAAS) pathway, although other hormonal imbalances may also be involved. Potassium is excreted in most of these disorders, but potassium loss does not cause hypertension. Vascular smooth muscle constriction and the inability of this tissue to relax do increase peripheral resistance, which can lead to hypertension, but this pathogenesis is not part of most hypertension-associated monogenic disorders.

PTS: 1 22. ANS: B

Mutations in many different genes can cause atrial fibrillation, which means that this arrhythmia has genetic heterogeneity. It is not related to long QT syndrome. Although it is seen more often in older adults, it can occur in children and can even be present at birth. The most common genetic cardiac disorder worldwide is familial hypertrophic cardiomyopathy, not atrial fibrillation.

PTS: 1 23. ANS: D

Long QT syndrome is a repolarization delay that makes the heart vulnerable to developing a potential lethal polymorphic ventricular tachycardia called torsade de pointes or TdP (meaning “twisting around a point”). It is not associated with atrial problems. Mitral valve prolapse is not

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

itself an arrhythmia.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 14: The Genetics of Cancer Multiple Choice Identify the choice that best completes the statement or answers the question.

1. A patient states that she has heard that the origin of most cancers is “genetic.” What is the

best response?

A. “The development of most cancers is predetermined and not affected by environmental factors.”

B. “Cancers arise in cells that have alterations in the genes.”

C. “Cancer is more common among males than females.”

D. “The majority of cancers are inherited.”

2. Which theory of carcinogenesis has the most support?

A. DNA damage, which permits overexpression of oncogenes B. RNA damage, which results in incomplete protein formation C. Autoantibodies, which attack specific “self” tissues and organs D. The failure of embryonic tissues to undergo normal differentiation 3. By which process does “promotion” assist in cancer development?

A. Inflicting mutations at specific sites on the exposed cell’s DNA B. Stimulating or enhancing cell division of cells damaged by a carcinogen C. Increasing the transformed cell’s capacity for error-free DNA repair D. Making cancer cells appear more normal and escaping immunosurveillance 4. How is progression different from metastasis?

A. Progression cannot occur unless the process of metastasis occurs first.

B. Metastasis occurs in both benign and malignant cells, whereas progression

is a feature that is unique to malignant cells.

C. Metastasis is dependent on gene mutations in suppressor genes, and progression

is dependent on gene mutations in oncogenes.

D. Progression involves continual gene changes in a cancer cell that enhance its

degree of malignancy, whereas metastasis is the ability of the cell to invade other tissues.

5. Which of the following benign tumors usually express aneuploidy?

A. Lipomas B. Leiomyomas C. Neurofibromas D. Neuroblastomas 6. How is a complete carcinogen different from an incomplete carcinogen?

A. Complete carcinogens damage oncogenes, and incomplete carcinogens

damage suppressor genes.

B. Complete carcinogens damage suppressor genes, and incomplete

carcinogens damage oncogenes.

C. Incomplete carcinogens are more likely to induce sporadic cancers.

D. Complete carcinogens are more likely to induce sporadic cancers.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

7. Which statement regarding the biology of cancer is always true?

A. Cancer cells arise from normal cells.

B. Testicular cancer is strongly associated with excessive masturbation.

C. When cancer cells are exposed to air, their growth rate becomes uncontrolled.

D. The biggest risk factor for cancer development is having a first-degree

relative with cancer.

8. Which of these qualities is common to cancer cells?

A. Apoptosis of damaged cancer cells occurs at a high rate.

B. Telomeres of cancer cells have pronounced shortening.

C. Their production of cell adhesion molecules is excessive.

D. They continue to grow even when surrounded by other cells.

9. How are malignant tumors different from benign tumors?

A. Malignant tumors grow by expansion, and benign tumors grow by invasion.

B. Malignant tumors lose plasma membranes, and benign tumors continue to

produce them.

C. Benign tumors retain parental cell functions, and malignant tumors lose

parental cell functions.

D. Benign tumors have totally normal features, and malignant tumors have

totally abnormal features.

10. Which feature is considered anaplastic?

A. Loss of a distinctive appearance B. Having a larger nuclear-to-cytoplasmic ratio C. Failure to undergo apoptosis at the appropriate time D. The ability to undergo mitosis when nutrition is poor 11. Which cancer type is associated with a 9;22 translocation t(9;22)?

A. Acute promyelocytic leukemia B. Acute lymphocytic leukemia C. Chronic lymphocytic leukemia D. Chronic myelogenous leukemia 12. By which process does “initiation” assist in cancer development?

A. Enhancing the cell division of cells damaged by a carcinogen B. Inflicting mutations at specific sites on the exposed cell’s DNA C. Increasing the transformed cell’s capacity for error-free DNA repair D. Making cancer cells appear more normal and escaping immunosurveillance 13. Which statement best describes the role of tumor suppressor genes in cancer development?

A. Tumor suppressor genes control or modify the activity of oncogenes, reducing

the risk for cancer development.

B. The presence of tumor suppressor genes increases the risk for gene damage by environmental carcinogens.

C. Tumor suppressor genes reduce/suppress immune function, increasing the risk for cancer development.

D. Tumor suppressor genes are a type of oncogene that is only active in germline cells

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

and tissues.

14. Which type of body tissue has the highest risk for cancer development?

A. Bone tissue because its absorption of radiation is cumulative B. Connective tissue that remains functional throughout life C. Brain tissue because it does not respond well to injury D. Any tissue that retains the ability to divide 15. What event occurring during the latency period of carcinogenesis is most likely to contribute

to cancer development?

A. Cellular apoptosis B. Error-free DNA repair C. Exposure to promoters D. Oncogene inactivation 16. What is the result of a mutation occurring in a suppressor gene?

A. Gain of a new function B. Loss of an existing function C. Increased “error-prone” DNA repair D. Increased unequal “crossing over” during meiosis I 17. Which statement regarding general cancer development is true?

A. The risk for cancer development increases with age.

B. Cancers usually develop in tissues that are missing a nucleus.

C. Children of older mothers have a greater risk for cancer development.

D. Most mutations leading to cancer development occur in structural genes.

18. How does an MSH2 gene mutation contribute to the development of colon cancer?

A. Suppressor gene function is enhanced.

B. DNA mutations are incorrectly repaired.

C. Trinucleotide repeat sequences are enhanced.

D. Drug resistance genes undergo amplification.

19. Why are people who have poor DNA repair mechanisms at greater risk for cancer

development?

A. Their cancers are usually resistant to chemotherapy.

B. They have sustained the initial “hit” in all cells and tissues.

C. Their somatic mutations are more likely to be permanent.

D. They have greater exposure to environmental carcinogens.

20. How does an acquired mutation in a somatic cell gene leading to cancer development affect a

person's ability to pass on a predisposition for that cancer type to his or her children?

A. The predisposition can only be passed on if the person with the somatic cell mutation is female.

B. The risk for predisposition is dependent upon which tissue type experienced the somatic mutation.

C. Multiple somatic mutations are required for passing on a predisposition to cancer development.

D. There is no risk of passing on a cancer predisposition from a somatic cell mutation.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

21. What is the function of a normal BRCA1 gene?

A. Enhances overall cell growth during puberty B. Directs the development of normal breast tissue C. Increases the expression of cytochrome P450 enzymes D. Suppresses the growth potential of a variety of oncogenes 22. Which feature is associated exclusively with sporadic cancer?

A. The cause is unknown.

B. It usually affects both bilateral organs.

C. It occurs at the same frequency within a kindred as in the general population.

D. It is more likely to occur in younger people with few environmental risks than in

older people.

23. What percentage of common cancers appears to be hereditary?

A. 1% to 3% B. 5% to 15% C. 20% to 25% D. About 35% 24. Which characteristic(s) is/are associated with an inherited predisposition for a cancer type?

A. Cancers tend to appear at an earlier age than do “sporadic” cancers.

B. These cancers are not picked up by routine cancer screening methods.

C. The carcinogenesis stage of “promotion” is not required for cancer development.

D. They are passed on only to the children of the same gender as the parent with

the predisposition.

25. Juliet tells a nurse that she has three aunts (two on her father’s side, ages 42 and 56, and one on

her mother’s side, age 62) who were diagnosed with breast cancer. She asks if she should have genetic testing. What should the nurse tell her?

A. “Your family history indicates a high risk, and you should definitely have genetic testing.”

B. “Because no men in your family are affected, it is not inherited cancer, so you don’t need mammograms or any special screening practices.”

C. “Because your aunts were older when they got breast cancer, it was probably sporadic, and you should just have regular mammograms like everyone else.”

D. “Your family history may indicate an increased risk for breast cancer, and a genetic counselor could help determine whether you could benefit from genetic testing.”

26. A 40-year-old man who has a mother who was diagnosed with breast cancer at age 45, a father

who was diagnosed with smoking-related lung cancer at age 55, a 33-year-old sister with breast cancer, and a 38-year-old sister with ovarian cancer asks if he should be concerned for his cancer risk. What is the best response?

A. “Your risk is not affected by this family history because most of the cancers arose in female sex–associated tissues.”

B. “You have two first-degree relatives and two second-degree relatives with cancer, which increases your general risk for cancer.”

C. “Your risk for breast cancer may be increased and requires more investigation;

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

however, your risk for lung cancer is not affected by this history.”

D. “Your risk for cancer is affected by your parents’ cancer development, and you should have genetic counseling on that basis; however, your sisters’ cancers have no bearing on your risk.”

27. Which statement about a “germline” mutation in either a cancer suppressor gene or an

oncogene is accurate?

A. Cancer risk is increased only in sex hormone–sensitive tissues.

B. The gene now has expressive potential but not penetrant potential.

C. Cancer risk increases, but additional mutations are required for cancer development.

D. A person inheriting such a mutation has a 100% risk for developing a specific cancer type.

28. A 22-year-old college student tells his nurse practitioner in the student health center that his

mother died of colon cancer at age 32. He asks if this could have an impact on his health. What is your best response?

A. “Yes, you need to have yearly stool tests for occult blood.”

B. “Yes, it would be good for you to talk with a genetics counselor.”

C. “No, because colon cancer is considered a type of sporadic cancer.”

D. “No, your risk would only be increased if your father had the colon cancer.”

Multiple Response Identify one or more choices that best complete the statement or answer the question.

29. Which personal factors indicate the possibility of a person having a BRCA1 or BRCA2 mutation?

Select all that apply.

A. The person has an adopted sister with ovarian cancer.

B. The person’s brother was diagnosed with breast cancer.

C. The person has always been 20 lb overweight as an adult.

D. The patient’s father died of pancreatic cancer at age 44.

E. The person is of Ashkenazi Jewish ethnicity.

F. The person’s 78-year-old grandmother was just diagnosed with breast cancer.

30. Mutations in which of the following genes are now known to greatly increase the risk for

developing breast cancer? Select all that apply.

A. APC B. CDH1 C. CHEK2 D. DCC E. PALB2 F. PTEN

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 14: The Genetics of Cancer Answer Section MULTIPLE CHOICE 1. ANS: B

Cancer development involves changes in the genes that regulate cell division. Although cancer is inherited from one cell generation to the next, most cancers and the risk for cancer development are not inherited from one’s parents; however, there are exceptions.

PTS: 1 2. ANS: A

Damage to the DNA can reduce the function of suppressor genes or enhance the function of oncogenes. Both types of damage reduce the regulation of cell growth, allowing growthpromoting oncogenes to be overexpressed.

PTS: 1 3. ANS: B

Initiation changes the structure or location of specific genes, usually oncogenes or tumor suppressor genes, altering their function. However, if cell division and growth do not occur, a cancer does not develop. Although promoters do not affect gene structure or location, they are growth-promoting factors that allow the mutations to be expressed as some form of abnormal cell growth.

PTS: 1 4. ANS: D

Metastasis only occurs in malignant cells. Progression also occurs only in malignant cells and involves further changes in genes that confer advantages to the cancer, causing it to lose more differentiated functions and become more highly malignant. Progression is a process that allows metastasis, not the other way around.

PTS: 1 5. ANS: A

Aneuploidy is having abnormal chromosome numbers and/or structures. Most benign tumors have normal chromosomes, a condition known as euploidy. Lipomas, which are benign fatty tumors, most often have structural rearrangements of chromosomes 6, 12, or 13. Leiomyomas and neurofibromas are benign tumors that are euploid. Although neuroblastoma is aneuploid, it is a malignant tumor.

PTS: 1 6. ANS: D

Complete carcinogens damage DNA for both suppressor genes and oncogenes and do not require additional exposure for promotion. They are most commonly associated with sporadic cancers rather than targeting genes that have already undergone a genetic change that increases predisposition to cancer development. Incomplete carcinogens do not have promoting capabilities and require that the initiated cells be exposed to a promoter before cancer can occur.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

7. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A key concept in cancer development is that cancer cells arise from normal cells through changes in genes that control cell division. (So, cancers are not alien beings from the planet carcinoma that just happened to invade the body.) Exposure to air does not change a cancer cell’s growth rate.

Masturbation is not related to testicular cancer development. Having a first-degree relative with cancer is only a risk factor for those cancers that have an inherited or familial pattern.

These make up only a very small percentage of cancers.

PTS: 1 8. ANS: D

A major feature of cancer cells is their lack of contact inhibition when membranes are touched on all sides by other cells. The telomeres do not shorten but are maintained by higher levels of telomerase. They stop producing cell adhesion molecules and become migratory. Apoptosis does not occur among cancer cells.

PTS: 1 9. ANS: C

Over time, malignant cells express fewer and fewer functions of the parental cells, whereas benign tumor cells retain parental cell functions. All cells must have plasma membranes for the life of the cell. Benign tumors grow by expansion, and malignant tumors are invasive. Although benign tumor cells retain many normal features and characteristics, they are not totally normal, or they would not form a tumor. The mere existence of benign tumor cells indicates that the strict regulation of growth has been overcome to some degree.

PTS: 1 10. ANS: A

Anaplasia is the loss of the specific differentiated appearance (morphology) of mature cells.

Having a large nuclear-to-cytoplasmic ratio is a feature of cancer cells but does not constitute anaplasia.

Loss of apoptosis and being able to divide when nutrition is poor are both functional changes rather than appearance changes.

PTS: 1 11. ANS: D

As cancer cells become more malignant, they usually become aneuploid, with gains or losses of whole chromosomes, chromosome breakage, and the structural rearrangements of chromosomes.

Some chromosomal rearrangements are unique to a cancer type and can be used to identify it as a specific cancer type. A chromosome rearrangement in which the ends of the q arms of chromosome 22 are translocated to the q arms of chromosome 9 results in the activation of a special tyrosine kinase that converts bone marrow cells into chronic myelogenous leukemia.

PTS: 1 12. ANS: B

The initial step or initiation in cancer development is mutating the DNA in such a way that either suppressor genes cannot perform their functions in cell growth regulation or oncogenes become resistant to suppressor gene control. Any substance or event that can damage DNA has the potential to mutate suppressor genes or oncogenes and is a carcinogen. Initiation is an irreversible event that can lead to cancer development if the cell’s mitotic ability remains intact.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 13. ANS: A

Suppressor gene regulation of cell division appears lost or defective in cancer cells, and oncogenes are then overexpressed, which leads to uncontrolled mitosis. They are not a type of oncogene, and they do not regulate immune function.

PTS: 1 14. ANS: D

Most cancers arise in tissues that have retained mitotic ability. The chance that these tissues will undergo spontaneous DNA replication error is greater than that for nondividing tissue. Also, most tissue types that remain mitotically active are located in body areas where tissue injury and exposure to environmental carcinogens are more common.

PTS: 1 15. ANS: C

Not all initiation events lead to cancer development. A major factor in having an initiated cell or tissue undergo malignant transformation is continuing exposure to promoters or promoting events during the latency period. DNA repair, apoptosis, and oncogene inactivation are more likely to lead to correction of the gene damage that occurred during initiation or suppression of cell division than they are to carcinogenesis.

PTS: 1 16. ANS: B

The most common way initiation leads to excessive oncogene expression and cancer development is by damaging any one of many suppressor genes. When a suppressor gene is damaged, it loses some or all of its function and can no longer express its products in the proper amounts to control oncogene expression.

PTS: 1 17. ANS: A

Although the risk for cancer development is higher in some families, most cancers are the result of exposure to environmental carcinogens. As a person ages, his or her exposures increase, and the damage is cumulative. Aging is the biggest risk factor for cancer development.

PTS: 1 18. ANS: B

The MSH2 gene is a DNA repair gene. When it is mutated, damaged DNA is either incorrectly repaired or not repaired. This problem can result in either mutations in suppressor genes or mutations in oncogenes. In either case, oncogene expression is increased, and cell growth is unregulated.

PTS: 1 19. ANS: C

DNA damage occurs on a daily basis for any person. Most of this damage is repaired by DNA repair mechanisms before cancer can develop. For those people who have poor DNA repair mechanisms, these constant and random DNA mutations do not get repaired and are permanent, often constituting initiation.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

20. ANS: D

Somatic mutations occur after conception in somatic tissues rather than germline tissues.

Because only germline tissues are passed to one’s children, mutations in somatic cells only affect the individual who developed them and are not passed on to the next human generation.

PTS: 1 21. ANS: D

The BRCA1 gene is a suppressor gene that is normally active in people of both genders. Its function is to control the expression of many oncogenes. It has no role in normal breast development, puberty, or enzyme function.

PTS: 1 22. ANS: C

Sporadic cancers usually occur as a result of environmental exposure or unknown factors and do not have any observable pattern of inheritance within a kindred, nor do they have an increased frequency within a kindred. At the cell level, mutations through carcinogenesis have occurred, disrupting the normal regulation of cell division, usually among somatic cells. These cancers are not present in higher-than-expected levels within three or more family generations. Although the cause of the cancer is not always known (making primary prevention difficult), there is no genetic predisposition for it.

PTS: 1 23. ANS: B

Overall, the percentage of cancers that occur as a result of the inheritance of a germline gene mutation ranges between 5% and 15%. Although this is a low percentage, people who have these mutations are at great risk for cancer development.

PTS: 1 24. ANS: A

Inherited cancers tend to appear at earlier ages because the process of initiation of one allele occurred at conception. So, these cells essentially have already gone halfway through initiation at conception and only require one additional allele mutation followed by promotion for a malignancy to occur. They are autosomal mutations and can be inherited by children of either gender and are detected by the same screening methods used for sporadic cancers.

PTS: 1 25. ANS: D

Juliet’s aunts on her father’s side were younger than 60 when they were affected. This would raise concern that there might be genetic risk factors in the family. The fact that they are on her father’s side makes no difference in genetic risk, and it is unclear whether or not Juliet should have genetic testing until she speaks with a genetics professional who can take a thorough family history and accurately discuss her risk.

PTS: 1 26. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

This man has four first-degree relatives with cancer (not just two). Those affected by breast cancer meet several criteria for inherited cancer. Even though the concerned patient is male, if he is BRCA1 positive, his risk for breast cancer (and some other types of cancers) would be increased. Most lung cancer is sporadic, with a strong environmental component, although a familial susceptibility can exist (but the genetic input is not known).

PTS: 1 27. ANS: C

There are a number of inherited germline mutations of suppressor genes or oncogenes that greatly increase the risk for cancer development. These cells essentially have already gone halfway through initiation at conception and only require one additional allele mutation followed by promotion for a malignancy to occur. This is why the cancer tends to appear at earlier ages than expected. Although its presence does not absolutely mean that the person will go on to develop cancer, the risk is much higher than for people who do not carry such a mutation.

PTS: 1 28. ANS: B

Death from colon cancer in early adulthood is a red flag for a genetic predisposition. However, this risk needs to be assessed by a genetics professional, not a general nurse practitioner.

Although checking stools for occult blood once yearly is a good idea, it is not sufficient surveillance if this student does have an increased risk for colon cancer. Colon cancer, either the sporadic type or one with a genetic predisposition, affects both genders equally.

PTS: 1 MULTIPLE RESPONSE 29. ANS: B, D, E

The indications of a possible BRCA1 or BRCA2 mutation in a family are as follows: o Cancers occurring at younger than 50 years of age o Breast cancer in male relatives o Breast cancer in both breasts o Triple-negative breast cancer (no estrogen, progesterone, or human epithelial growth receptors [HER2] on the surface of the cellular membrane) o Presence of a second primary cancer in the same patient o Ashkenazi Jewish ethnicity o Presence of family members with both breast and ovarian cancers o History of pancreatic cancer in any family member o Confirmation of a BRCA1/2 genetic mutation in the family PTS: 1 30. ANS: B, C, E, F

Along with BRCA1 and BRAC2, these four genes are associated with an increased risk for breast cancer development. APC is associated with an increased risk for colon cancer, stomach cancer, and pancreatic cancer. DCC is associated with an increased risk for colorectal cancers.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 15: Genetic Contributions to Psychiatric and Behavioral Disorders Multiple Choice Identify the choice that best completes the statement or answers the question.

1. What is meant by the phrase “behavioral genetics”?

A. Identification of the single gene that corresponds to a specific undesired behavior B. An area of research aimed at controlling behavior by genetic manipulation C. The study of the behavior of genes in diverse settings D. The examination of gene variants that influence behaviors 2. Which statement regarding behavioral genetics is accurate?

A. Genes and gene products have been discovered that directly control behavior.

B. A genetic predisposition toward a specific behavior can be modified by

altering environmental influences.

C. The genotypes and phenotypes of behavioral problems or deviations

follow Mendelian autosomal-recessive inheritance patterns.

D. The genetic susceptibility to or predisposition toward a behavioral

disorder requires the trigger of an infectious disease for expression.

3. Why is direct-to-consumer marketing of diagnostic tests for mental illness and behavioral

problems a dangerous practice?

A. Because the tests are expensive and not ordered by a health-care provider, they are not covered by insurance.

B. The tests are not predictive of which psychiatric medications are most likely to be beneficial to a specific individual.

C. People using such tests may not receive professional counseling for interpretation of results and accurate risk assessment.

D. These tests are often misused by employers to support the dismissal of employees who express opinions that are different from those of the company.

4. Which statement about attention deficit hyperactivity disorder (ADHD) is true?

A. The problem is more common among girls whose fathers also had the disorder.

B. By the time a person with ADHD reaches adulthood, all symptoms have resolved.

C. It is a complex disorder caused by gene variants along with

environmental contributions.

D. Symptoms of the disorder are ameliorated when the person sleeps more soundly

as a result of pharmacologic intervention.

5. What is the best description of the genetic contribution to the onset of autism?

A. Autism spectrum is caused by a single-gene mutation in most, but not all, cases.

B. Exposure to a teratogen can be a cause of autism spectrum in many cases.

C. Known causes of autism spectrum include copy-number variants and

chromosomal problems.

D. Autism spectrum disorders have a much stronger environmental input than

genetic input to expression of the phenotype.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

6. Which condition provides the greatest support for a strong genetic contribution to autism

spectrum disorders (ASDs)?

A. Concordance among dizygotic twins is 10%.

B. Males are affected four times more frequently than females.

C. People with some neurologic disorders also have features of ASDs. D. ASDs are more likely to occur in children of parents who are over 35 at the time of conception.

7. Which assessment finding in a 3-year-old child increases the suspicion for a possible diagnosis

of autism spectrum disorder?

A. Language skills are delayed.

B. The child is very near-sighted.

C. Height is below the 5th percentile.

D. The child does not sit still during the assessment interview.

8. Why is pharmacogenetics/pharmacogenomics of particular interest in treating patients

with psychiatric/mental health problems?

A. Psychiatric medications may be effective in only a small group of patients.

B. Genetics restricts patients to only one drug in each classification.

C. Most psychiatric illnesses are single-gene disorders with predictable drug responses.

D. Psychotropic medications have few side effects.

9. Which condition or behavior suggests attention deficit hyperactivity disorder (ADHD)?

A. A 5-year-old who does not want to go to bed at 8:00 p.m. and begs to be

allowed to stay up later B. An 8-year-old who can hop on one foot but cannot ride a bicycle without

training wheels C. A 9-year-old who does not focus on a favorite story or game for longer than

3 minutes D. A 10-year-old who prefers to play basketball rather than chess 10. Why are single-nucleotide polymorphisms (SNPs), which have been associated with

schizophrenia and bipolar disorder, currently not used to diagnose either of these disorders?

A. The changes in gene activity caused by the SNPs is not sufficient to cause disease expression.

B. SNPs are associated with nicotine dependence but not with schizophrenia or bipolar disorder.

C. The SNPs are too small to result in either a frameshift mutation or a point mutation.

D. These SNP-induced gene changes cannot be altered by pharmacologic therapy.

11. Schizophrenia is reported to have a heritability estimate as high as 80%. What does this mean?

A. If a parent has schizophrenia, each child has an 80% risk of getting schizophrenia.

B. If we look at a population, 80% of the risk for schizophrenia comes from genetics.

C. If your patient’s sibling has schizophrenia, her risk of getting schizophrenia

is 80%.

D. Genetics contributes 20% to the risk of schizophrenia.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

12. At what point in human development is the phenotype of schizophrenia most likely to manifest?

A. Between birth and 10 years B. At the onset of puberty C. At adolescence or early adulthood D. With the onset of the physical decline associated with older age 13. An adult daughter whose mother has bipolar disorder is concerned that she may also have the

disorder because she engages in all of the following behaviors. Which behavior should be explored further as a manifestation of bipolar disorder?

A. Crying whenever she looks at a picture of a close friend who died 3 months ago B. Refusing to allow her husband to participate in general housekeeping chores C. Having a strong sense of sleepiness daily in the middle of the afternoon D. Making frequent purchases of expensive clothing that she never wears 14. A 48-year-old woman reports that her 50-year-old brother has severe bipolar disorder, and she is

concerned that her two teenage daughters may be predisposed to the problem. She has no manic or depressive symptoms. How should you respond to her concerns?

A. “If your household is stable and loving, your daughter’s increased risk can be minimized.”

B. “If your husband also has no symptoms of the disorder, your daughters have the same risk as the general population.”

C. “Because the heritability of bipolar disorder is 50%, it is likely that only one of your two daughters will develop the disorder.”

D. “Because your brother is a blood relative, your daughters have a risk for bipolar disorder that is two to three times greater compared with the general population.”

15. What is the heritability estimate for addiction to alcohol in both males and females?

A. 10% to 20% B. 20% to 40% C. 50% to 60% D. 60% to 80%

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 15: Genetic Contributions to Psychiatric and Behavioral Disorders Answer Section MULTIPLE CHOICE 1. ANS: D

Behavioral genetics is the area of study that focuses on the gene variants that influence behaviors.

Behaviors are unlikely to be caused by single-gene problems, and it is unlikely that behaviors will be controlled by genetic manipulation.

PTS: 1 2. ANS: B

The interaction of genes, environments, and behaviors is very complex, and a genetic predisposition for a certain behavior may be altered over time with changes in diet and parenting.

Genes and/or gene products may influence behavior but do not directly control it. No mental illness or behavior disorder always follows autosomal-recessive inheritance patterns. At the present time, although an environmental trigger has been speculated to result in the expression of a predisposition to a behavioral disorder, no trigger has been identified.

PTS: 1 3. ANS: C

Although it is true that the tests can be expensive and not covered by insurance, and they are not predictive of which drugs would be most beneficial, the biggest danger is that the person testing at home does not have resources to interpret results and assess risk. The outcomes of such testing could have a devastating effect on a person who tested positive but did not have appropriate access to a counselor for interpretation of this result. Testing of any person for any reason without his or her knowledge and assent is illegal.

PTS: 1 4. ANS: C

Genome-wide association studies (GWASs) provide substantial support that ADHD is a complex (multifactorial) disorder caused by the actions of many gene variants along with environmental contributions. GWASs have found regions on more than 10 chromosomes that are associated with ADHD in some people. The disorder is much more prevalent among males and is not dependent on which parent may also have had the problem. More sleep does not ameliorate the problem. Although many adults who had ADHD as a child have fewer symptoms in adulthood, about 2% to 4% retain most symptoms into adulthood.

PTS: 1 5. ANS: C

Autism is a diverse disorder associated with a variety of genetic and environmental issues.

Genetic causes of autism include chromosomal problems and copy-number variants. Exposure to teratogens is the cause in a small number of cases, and single-gene causation is rare. Between 1% and 3% of people with autism have fragile X syndrome.

PTS: 1 6. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Dizygotic (fraternal) twin concordance is about 10%. This may not seem very high at first glance, but it is about 100 times higher than the risk of autism in someone from the general population.

That is a huge difference and indicates that autism is probably highly heritable. The increased incidence of ASD among males compared with females suggests a possible biologic contribution but not necessarily a genetic contribution. Because some features of ASD indicate neurological problems or deficits, people who have known neurologic problems may share these characteristics, but the origin can be different. Advanced parental age has not been consistently associated with ASD. PTS: 1 7. ANS: A

Delay of language skills in early childhood is a hallmark of autism spectrum disorders (ASDs), although they also are associated with other problems. Visual disorders do not consistently accompany ASDs, and neither does short stature. Most 3-year-olds have difficulty sitting still during any type of interview.

PTS: 1 8. ANS: A

Many psychiatric patients are prescribed a variety of drugs before they find one that works well for them. PGx promises to help clinicians find the right drug for the right patient by drugresponse genetic testing. Most psychiatric diseases are not single-gene problems, and psychotropic medications tend to have significant numbers of side effects.

PTS: 1 9. ANS: C

The hallmark of ADHD is not just being physically active, which many healthy children are, but not being able to focus on anything, even those events or activities the child really enjoys.

PTS: 1 10. ANS: A

Schizophrenia, bipolar disorder, and nicotine dependence have been found to have singlenucleotide polymorphisms (SNPs) that alter a single DNA base. However, so far, these SNPs have not been shown to be significant enough to bring about a psychiatric disorder. Instead, they increase or decrease the risk by a small fraction, much less than twofold, so testing one of these does not warrant developing it as a diagnostic test. SNPs do cause a point mutation and can cause a frameshift mutation. At this time, no gene changes in psychiatric or behavioral disorders can be altered by pharmacologic therapy.

PTS: 1 11. ANS: B

The heritability estimate for schizophrenia is about 80%, which refers to the cause of the variation of the phenotype within a population and not causes within a given family. Even though the heritability estimate is 80%, that does not mean that if a mother is schizophrenic that each of her children has an 80% risk of being schizophrenic. It does mean that examination of the variations in phenotype in specific populations demonstrates that about 80% can be attributed to genetics and about 20% can be attributed to environment.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

12. ANS: C

Environmental and genetic factors both have roles in the development of schizophrenia, although the exact nature of the genetic influence has yet to be determined. Copy-number variants appear to play the most important role in risk for the disorder. The most common age of onset is early adulthood.

Monozygotic (identical) twin concordance is between 40% and 50%. It is likely that there are epigenetic contributions as well. There is no laboratory test that can determine if someone has schizophrenia, and sometimes the variability in the phenotype makes diagnosis based on clinical observation and self-report difficult.

PTS: 1 13. ANS: D

Bipolar disorder (BPD) is characterized by episodes of mania and depression that usually follow each other in cycles. The manic episodes consist of at least 1 week of elated or irritable mood that is accompanied by racing thoughts and highly pressured speech. The person is easily distracted and agitated and may engage in hypersexuality and out-of-control spending. Crying when reminded of the recent death of a loved one is not true depression, and neither is afternoon sleepiness. Insisting on performing all housekeeping chores more likely indicates an issue with role identification or association than manic behavior.

PTS: 1 14. ANS: B

The risk for the disorder increases when an affected person is a first-degree relative. If neither parent has the disorder, this risk for the two daughters does not increase above that for the general population because an uncle is a second-degree relative. The heritability estimate suggests that the risk is about even between genetic and environmental factors.

PTS: 1 15. ANS: C

It appears that nearly equal genetic and environmental influences contribute to addiction to alcohol. The heritability estimate is between 50% and 60%.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 16: Genetic and Genomic Testing Multiple Choice Identify the choice that best completes the statement or answers the question.

1. How is cell-free DNA (cdDNA) testing being used clinically?

A. To determine a person’s risk for developing breast cancer B. As an inexpensive alternative to standard paternity testing C. As a less invasive way to determine the characteristics of cancer cells D. As a way to determine whether a specific suspect has committed a rape 2. Genetic testing that examines an asymptomatic person’s DNA sequence, looking for mutations

that increase a person’s susceptibility to a disease, is an example of which type of testing?

A. Diagnostic testing B. Predispositional testing C. Presymptomatic testing D. Cytogenetic testing 3. Which tissue is most likely to provide an adequate DNA sample for genetic testing?

A. Distal ends of hair shafts B. Mature red blood cells C. Nasal epithelial cells D. A mummy's tooth 4. What type of genetic test provides information about an asymptomatic person’s risk for

having a child with a specific autosomal-recessive disorder in the future?

A. Carrier test B. Diagnostic test C. Newborn screening D. Predictive test 5. Under which condition can preimplantation genetic testing be performed?

A. During in vitro fertilization B. Between 4 to 6 weeks’ gestation C. When paternity is unknown D. When an ultrasound indicates the fetus has a structural abnormality 6. Genetic testing of an asymptomatic woman whose sister is positive for the mutation that is

responsible for an autosomal-dominant disorder that has a 70% penetrance rate would fall into which testing category?

A. Diagnostic predisposition B. Diagnostic presymptomatic C. Predictive predisposition D. Predictive presymptomatic 7. Which type of genetic testing is the most sensitive method for detecting any mutation in a

specific gene?

A. Immunohistochemistry

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. Direct DNA sequencing C. Banded chromosomal analysis D. Fluorescence in situ hybridization 8. Which type of genetic testing examines a person’s chromosomes for variations in number or

structure?

A. Cytogenetic testing B. Preimplantation testing C. Predictive testing D. Electropherogram testing 9. How is the polymerase chain reaction (PCR) helpful in the process of genetic testing?

A. Preserving genetic material within a sample so that more sophisticated tests can be

performed as they become available with future technologic advancements B. Separating exons from introns and establishing which DNA strand is the “sense”

strand C. Determining whether a small amount of DNA is paternal or maternal in origin D. Increasing the amount of DNA being tested, thus promoting accuracy 10. What is the major disadvantage to the fluorescence in situ hybridization (FISH) method of genetic

testing?

A. Single-nucleotide mutations cannot be detected.

B. Depending on the skill of the technician, the rate of false positives is high.

C. The sample for testing must be obtained from living tissue rather than from a preserved specimen.

D. Results are not available quickly because cells must first be grown in culture before testing can be performed.

11. When is fluorescence in situ hybridization (FISH) most likely to be used?

A. To find a gene associated with a particular disease B. To determine whether a stillborn infant has trisomy 13 C. To determine the carrier status of a child whose sister has cystic fibrosis D. To establish a diagnosis for a possible adult-onset single-gene disorder 12. A woman whose sister tested positive for a specific mutation in the BRCA1 gene, which increases

the risk for breast and ovarian cancer, is found not to have that mutation but does have a mutation of unknown significance near the known mutation site. How should this woman be counseled?

A. She should be informed that her risk for breast cancer is greater than that of the general population but not as great as her sister’s risk.

B. She should be informed that because she does not have the mutation, her risk for breast cancer is not greater than that of the general population.

C. She should be informed that she does not have the specific mutation but that because another mutation is present, she should be vigilant about screening.

D. She should be informed of her gene-mutation status and be presented with all the available prophylaxis options and reconstruction options.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

13. A patient is 34 years old and concerned about possibly being a carrier for HNPCC because his

father died of colon cancer at 39, his father’s sister died of colon cancer at age 41, and his brother (aged 37) now has colon cancer. The brother’s testing is negative for all the known mutations associated with inherited forms of colon cancer. How should this patient be counseled about his risk for colon cancer?

A. Explain that because the brother with cancer is negative for these gene mutations, this cancer is most likely sporadic, and his risk is the same as that for the general population.

B. Explain that he could benefit from testing even though his brother is negative for these mutations because reduced penetrance might account for his negative status.

C. Explain that testing for him would be of no benefit because of the current test limitations but that his family history does place him at high risk.

D. Explain that his risk is not related to his brother’s diagnosis because he did not inherit any genes from him, but because his father is a first-degree relative, testing should be considered.

Multiple Response Identify one or more choices that best complete the statement or answer the question.

14. Which genetic-based disorders can be initially identified by the most common types of

newborn blood screening? Select all that apply.

A. Congenital hypothyroidism B. Cystic fibrosis C. Galactosemia D. Hereditary hemochromatosis E. Phenylketonuria F. Sickle cell disease

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 16: Genetic and Genomic Testing Answer Section MULTIPLE CHOICE 1. ANS: C

Cell-free DNA is small pieces of nongenomic DNA circulating in the blood that can be collected from plasma or urine. These very small pieces of circulating cfDNA originate from the fetus in pregnant women or tumors in cancer patients. Because this technology is noninvasive, it offers a reduced risk of complications compared with the currently used invasive technologies. It can be used for some types of fetal testing. In oncology, cfDNA is being used to identify information about the tumor, including the genetics, amount of burden, mechanisms of progression, and drug resistance.

Another advantage of this developing technology is that it costs much less.

PTS: 1 2. ANS: B

Predispositional testing is done when having a gene variant increases the likelihood that a person will develop a genetic disease, but that does not mean that the person is certain to get it.

Diagnostic testing is done to confirm or rule out a particular diagnosis in a symptomatic person.

Presymptomatic testing indicates that the individual will develop the disease he or she was tested for at some point in the future. Cytogenetic testing is the examination of chromosomes for markers, structural rearrangements, and abnormalities of number.

PTS: 1 3. ANS: C

Any tissue with a true nucleus, such as nasal epithelial cells, can provide DNA for genetic testing.

The distal ends of hair shafts do not contain cells with a nucleus; only the hair root does. Mature red blood cells have no nucleus (and no mitochondria). A mummy’s tooth would have enamel with no cells, and the tooth pulp would be long gone.

PTS: 1 4. ANS: A

A carrier test indicates if a person is heterozygous for a mutation that can cause an autosomal-recessive disease. A predictive test provides information about an asymptomatic person’s risk of getting a genetic disease in the future. Newborn screening provides information about whether or not a baby is likely to have the disease being tested for, and a diagnostic test indicates whether or not a person currently has a given disease.

PTS: 1 5. ANS: A

Preimplantation genetic diagnosis is a process only done in conjunction with in vitro fertilization because it is performed on early embryonic cells that are in a petri dish and have never been in the mother’s body. A group of embryos generated by in vitro fertilization is removed from the eight-cell blastocyst. Cells from each embryo can be tested to find gene variants causing singlegene disorders or chromosomal problems, or to determine sex. One or two “healthy” embryos are then selected for implantation.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 6. ANS: C

The test is predictive rather than diagnostic because the woman does not have any symptoms of the disorder. So, the test is not being performed to confirm or rule out a disorder diagnosis. The test is also considered “predisposition” testing rather than “presymptomatic” testing because the disorder has a 70% penetrance rate. This means that even if the woman is positive, she has a 30% chance of never developing symptoms but has a genetic predisposition to the disorder.

PTS: 1 7. ANS: B

DNA sequencing is the analysis of the bases in a stretch of DNA. This type of testing is considered the most sensitive because it can even detect a difference in one base within a sequence. It is also expensive and is not used unless no other type of test can provide appropriate critical and accurate information.

PTS: 1 8. ANS: A

The type of test that examines chromosomes is cytogenetic testing. Preimplantation testing examines an embryo generated by in vitro fertilization before it is implanted. Predictive testing tests an asymptomatic person to determine future genetic risk. An electropherogram is a graphic illustration of the nucleotide sequence in a sample.

PTS: 1 9. ANS: D

The polymerase chain reaction precisely expands or amplifies a specific segment DNA so that more of it is available for testing. PCR itself does not test the DNA directly for anything, cannot itself determine whether DNA or a gene is paternal or maternal, and does not preserve specimens.

However, because of amplification, PCR can help ensure that enough DNA is present to reliably perform an accurate analysis. This amplification may allow more of a specimen to be preserved but does not actually preserve the specimen.

PTS: 1 10. ANS: A

FISH examines portions of chromosomes, not the sequence of individual nucleotides. The test can determine the number of chromosome copies, areas of duplication, or deletions; however, even with microdeletions, the test cannot detect changes in segments that are smaller than 100 nucleotides.

FISH can be performed on preserved tissue as well as living tissue specimens. It is an accurate test with great specificity and a low risk for either false positives or false negatives.

PTS: 1 11. ANS: B

FISH is often used to identify chromosomal variations in number or structure and to identify deletions or duplications. The goal is to determine the presence or absence of a chromosomal type of genetic variation.

PTS: 1 12. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Inconclusive test results can be confusing to patients. These results mean that a known mutation has not been identified, but some gene variant has been found. It is unknown at this time whether or not it can result in increased cancer risk. This patient should continue with recommended screening guidelines based on her family history.

PTS: 1 13. ANS: C

Further testing for common mutations would be of no benefit to this patient. His risk is not known but should be considered high on the basis of family history. Although some of the more common single-gene mutations responsible for serious health problems have been identified, not all have been identified. In addition, some mutations may be specific for a family and not present in the general population. When a disorder can be identified by family history as following autosomal-dominant inheritance patterns, but no known mutation has been identified, all firstdegree relatives should be managed in terms of surveillance, screening, and prophylaxis as if they had a known mutation.

PTS: 1 MULTIPLE RESPONSE 14. ANS: A, C, E, F

Newborn screening is done to identify those infants at high risk of a variety of disorders for which immediate treatment or intervention is available. The tests are usually biochemical rather than gene based, but results can indicate the likelihood of a genetic disorder being present.

Although testing varies somewhat by state, every state tests for these autosomal-recessive disorders: phenylketonuria, congenital hypothyroidism, galactosemia, and sickle cell disease.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 17: Assessing Genomic Variation in Drug Response Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which of the following is a goal of pharmacogenetics?

A. Producing “blockbuster drugs” that will work equally well for everyone B. Bringing down the cost of pharmaceutical manufacturing C. Developing drugs that will treat very rare diseases D. Reducing adverse reactions 2. Which processes of drug response are most subject to genetic variation?

A. Drug dissolution in body fluids and drug binding to plasma proteins B. Rates of drug movement into and through the gastrointestinal tract C. Drug activation or deactivation and duration of drug actions D. Drug binding with receptors and drug blocking of receptors 3. Which statement about agonist and antagonist drugs is true?

A. Both drugs types must interact with receptors to produce their intended responses.

B. The primary target tissues for these types of drugs are invading bacteria

and viruses.

C. Antagonists drugs produce only intended responses, and agonist drugs

produce both intended responses and side effects.

D. These types of drugs are less likely to cause allergic responses than drugs that

are neither agonists nor antagonists.

4. What is the expected heart-rate response when a patient is taking a drug that is an

adrenaline antagonist?

A. Heart rate is unchanged.

B. Heart rate decreases.

C. Heart rate increases.

D. Heart rate is irregular.

5. How do genetic/genomic issues influence individual variation in the response to a specific drug?

A. Polymorphisms of genes encoding metabolizing enzymes B. Age-related loss of alleles within the cells lining the intestinal tract C. Single-gene disorders that reduce the function of the kidneys or the liver D. Genetic-based behavior problems that promote poor adherence to prescribed

drug regimens 6. Which organ has the greatest concentration of cytochrome P (CYP) 450 enzymes?

A. Stomach B. Kidney C. Brain D. Liver 7. How are the actions of naturally occurring estrogen changed when a patient is taking a drug

that is an estrogen agonist?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. Actions are increased.

B. Actions are decreased.

C. Actions are eliminated.

D. Actions are unchanged.

8. Which type of parent compound must undergo first-phase metabolism to produce the

intended response after it enters the body?

A. A prodrug B. An active metabolite C. An inactive metabolite D. A sustained-released drug 9. What would be the patient’s response to a normal drug dose that, because of a genetic variation in

an enzyme that prepares the drug for elimination, results in a blood drug level that is below the minimum effective concentration (MEC)?

A. Drug entry exceeds drug elimination.

B. The risk for toxic side effects is increased.

C. The intended response fails to be produced.

D. The drug’s duration of action is longer than expected.

10. A patient has a very high concentration of insulin receptors on cells that require insulin for

glucose to enter. How should insulin dosages be adjusted for this patient to have blood glucose levels within the normal range?

A. Insulin dosages should be given less frequently because the drug will remain bound to receptors longer.

B. Insulin dosages should be decreased because the drug will exert its actions at lower concentrations.

C. Insulin dosages should be given more frequently because the drug will be eliminated at a faster rate.

D. Insulin dosages should be increased because cells will be less sensitive to the presence of active insulin.

11. Which condition or factor improves the initial bioavailability of a drug agonist that has

extensive first-pass loss as a result of the patient’s enhanced liver enzyme activity?

A. Increasing the patient’s fluid intake B. Co-administering the drug with an antagonist C. Administering the drug by the intravenous route D. Crushing the oral form of the drug before administration 12. A patient with a fractured elbow in the emergency department states that he needs morphine for

pain rather than codeine because the last time he had a painful injury, codeine was not effective in managing his pain. What is the nurse’s best response or action?

A. Ask the patient how much alcohol he ingests daily.

B. Communicate this information to the admitting physician.

C. Alert the health-care provider that this patient is “drug-seeking.”

D. Reassure the patient that he will receive progressively higher dosages of codeine until his pain is controlled.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

13. A patient is identified as an ultra-rapid metabolizer for drugs that are prepared for elimination

by CYP2D6. What effect will this have on the patient’s ability to benefit from any active drug that is metabolized by this enzyme?

A. Intended responses increase, whereas side effects decrease.

B. Intended responses decrease, whereas side effects increase.

C. Both intended responses and side effects increase.

D. Both intended responses and side effects decrease.

14. What is the most likely outcome for a patient with glucose-6-phosphate dehydrogenase

(G6PD) deficiency when he or she takes aspirin or an antimalarial drug?

A. Rapid elimination of the drug with no benefit B. Liver damage or failure C. Hemolytic anemia D. Fluid retention 15. An Asian American man and his Caucasian wife are both taking warfarin (Coumadin) daily

because of atrial fibrillation. The husband asks why he is prescribed a much-smaller-thanaverage dose of the drug to keep his international normalized ratio (INR) at 2.0, whereas his wife takes the average dose, even though he is taller and heavier than she is. What is the nurse’s best response?

A. “Body size is not important for warfarin, but gender differences are because testosterone improves its action.”

B. “You are probably anemic, which would reduce your ability to form blood clots, so your doses can be lower.”

C. “Many Asian Americans do not break down warfarin as fast as Caucasians, so the drug is more effective at lower dosages.”

D. “Caucasians have higher levels of the enzyme that breaks down warfarin, requiring higher dosages for the same effect on INR.” 16. Your patient has been identified as a poor metabolizer of a drug that has just been ordered, and

the drug is formulated as an active compound. What will be the most likely result if you give this active drug at the standard dose?

A. No therapeutic response B. Increased adverse reactions and possible toxicity C. Therapeutic response as expected in the general population D. Drug inactivation occurs more rapidly, and the therapeutic response is limited.

17. You are caring for a child with acute lymphoblastic leukemia. She has been genotyped and is

homozygous for a TPMT polymorphism, producing very little of the enzyme needed for this drug's metabolism. How would you expect this to affect dosing of the drug 6-mercaptopurine?

A. This child should receive only a small fraction of the standard dose.

B. This child should receive the drug intravenously rather than orally.

C. This child should receive higher doses than the standard dose.

D. This child should receive the standard dose.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 17: Assessing Genomic Variation in Drug Response Answer Section MULTIPLE CHOICE 1. ANS: D

The pharmacogenetic variations in individual drug response are based on polymorphisms in genes coding for metabolizing enzymes, transporters, and receptors. The clinical responses to these differences can range from life-threatening adverse reactions to a complete lack of therapeutic effect. About 20% of drugs produce adverse reactions that were unknown when the drugs first came to market, and adverse drug reactions are considered one of the leading causes of death. A main goal of pharmacogenomics is the reduction of adverse reactions to medications.

PTS: 1 2. ANS: C

Drug activation and deactivation are part of the pharmacodynamics, which involves drug metabolism and elimination. The enzymes involved in these processes influence the duration of drug actions. These responses are controlled by specific genes and thus are most subject to genetic variation. Drug dissolution is related to the chemical structure of the drug. Binding with plasma proteins is related to the amount of plasma proteins and the structure of the drug rather than genetic influences. The rate of drug movement within the gastrointestinal tract is variable, but the variation is physical rather than genetic. Although the number of receptors is genetically controlled, the ability of a drug to bind with a receptor functionally or nonfunctionally (blocking it) is mostly related to the structure of the drug.

PTS: 1 3. ANS: A

These drugs must interact with cell receptors to produce an effect. Some drugs activate receptors, working as “agonists” that increase the cells’ responses and activities. Other drugs block receptors, working as antagonists, which then reduces the effects of naturally occurring body substances on the cells’ responses and activities.

PTS: 1 4. ANS: B

Adrenaline (epinephrine) is made in the body and normally binds to heart receptors and increases the heart rate. A drug that is an adrenaline antagonist binds to heart cell receptors nonfunctionally and blocks naturally occurring adrenaline from binding to these cells. As a result, heart rate decreases (and often becomes more regular).

PTS: 1 5. ANS: A

The variations in individual drug response are based on polymorphisms in genes coding for metabolizing enzymes, transporters, and receptors. These polymorphisms are very common, whereas single-gene disorders of any type are not common. Loss of alleles has not been associated with the aging process. Behavioral problems are poorly understood, as are the genetic bases of specific behaviors.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 6. ANS: D

Although CYP-450 enzymes are present in many body tissues/cells, the liver has the greatest concentration. This organ is the tissue most actively involved in drug metabolism.

PTS: 1 7. ANS: A

An agonist drug has the same effects as the naturally occurring drug. So, taking a drug that is an estrogen agonist increases the effects of the patient’s naturally occurring estrogen.

PTS: 1 8. ANS: A

A prodrug is an inactive parent compound that must undergo first-phase metabolism to become an active drug that is capable of inducing an intended response. A sustained-release drug can be an active compound without first-phase metabolism, but the drug formulation allows slow and continuous release of the drug from its tablet or capsule. A parent compound is the actual drug as it enters the body and is not a metabolite.

PTS: 1 9. ANS: C

In order for a drug to produce its intended response, its blood levels must reach the minimum effective concentration. When enzymes responsible for metabolizing and preparing drugs for elimination are more active as a result of a genetic variation, sensitive drugs are eliminated more rapidly, and a “normal” dose may not be high enough to allow the drug to be retained long enough to reach the MEC. PTS: 1 10. ANS: B

For insulin or any drug that requires interaction with a receptor to agonize a cell and induce its intended response, an increase in the number of receptors specific to the drug increases the cell’s sensitivity to the drug. A smaller dose is needed to produce the response of a lower blood glucose level. Normal dosages could easily cause severe hypoglycemia.

PTS: 1 11. ANS: C

Bioavailability is the percentage of drug present in the bloodstream. When a drug is administered intravenously, it is all (100%) bioavailable initially. Increasing fluid intake does not affect bioavailability regardless of whether the fluid is taken orally or administered intravenously.

Administering an antagonist along with a drug agonist would only cancel out or negate the agonist action, not improve the bioavailability. Crushing an oral drug does not prevent first-pass loss or improve bioavailability.

PTS: 1 12. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A significant number of people have low or absent levels of the enzyme that converts the prodrug codeine to the active compound morphine at the cellular level. These people have no pain relief from any dose, even high doses, of codeine but do achieve pain control with usual dosages of morphine.

In the past, such individuals have been labeled drug-seekers. Although individuals who consume large amounts of alcohol often metabolize opioids very quickly and need higher dosages for pain control, this factor affects both codeine and morphine and is not selective just for codeine.

PTS: 1 13. ANS: D

A person who is an “ultra-metabolizer” for an active drug breaks down and eliminates the drug very quickly, often not achieving the minimum effective blood concentration. As a result, both the beneficial intended responses and the risks for side effects are greatly reduced at normal drug dosages.

PTS: 1 14. ANS: C

G6PD is an enzyme needed for red blood cells (erythrocytes) to maintain integrity during oxidative stress. Certain drugs, such as aspirin, antimalarials, probenecid, and vitamin K, increase the level of oxidative stress on red blood cells. With a deficiency of G6PD, the increased oxidative stress caused by these drugs leads to rapid destruction of red blood cells that is greater than the rate at which they can be produced by the bone marrow, resulting in hemolytic anemia.

PTS: 1 15. ANS: C

Polymorphisms in the CYP2C9 metabolizing enzyme are common in Asians and Asian Americans. As a result, the drug warfarin remains in the body longer. These individuals require lower dosages to maintain the prescribed INR and avoid serious bleeding. This response is not related to gender or body size.

PTS: 1 16. ANS: B

Active-compound drugs work as soon as they enter the bloodstream, without further processing.

They must undergo metabolism to be made ready for elimination. In people who are poor metabolizers, the blood drug levels are higher, and the drug remains active much longer. As a result, the patient is at higher risk for continued drug actions, including adverse effects and toxic responses. This person would need lower drug doses given less frequently.

PTS: 1 17. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Drugs such as azathioprine (Imuran), mercaptopurine, and thioguanine used to treat childhood leukemias, rheumatoid arthritis, or inflammatory bowel disease require metabolism by the enzyme thiopurine methyltransferase (TPMT). The activity of TPMT is trimodal, meaning three different levels of enzyme activity exist among groups of people. About 90% of people have high TPMT activity. They have faster drug metabolism, which results in lower exposure of leukemic cells to active thiopurines. About 10% of people have intermediate activity, and a very small fraction of people (about 0.3%) have low activity. Those who have low TPMT activity are homozygous for variations in the gene coding for TPMT, so they produce nonfunctional protein.

People with low TPMT activity risk myelosuppression, secondary cancers, and possibly fatal toxicity when they are given chemotherapeutic agents requiring metabolism by TPMT. They must either have the dose of their chemotherapeutic agents reduced eight- to tenfold or be placed on drugs that do not use TPMT for metabolism and elimination.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 18: Health Professionals and Genomic Care Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Which action reflects promotion of genomic care as part of comprehensive health care?

A. Calculating the odds ratios and recurrence risks of common complex

health problems for all people B. Ensuring that genomic issues potentially influencing a person’s health

are incorporated into routine care C. Encouraging all people to undergo genetic testing as part of a proactive

health screening and prevention program D. Instructing patients who undergo genetic testing about their obligation to

inform family members regarding test results 2. Why are general physicians and surgeons not considered to be genetics professionals?

A. The typical patients seen by these health-care providers have acute

conditions rather than chronic conditions, and thus genetic influence is irrelevant.

B. Most of today’s physicians and surgeons were educated before completion of the human genome research project.

C. These health-care professionals have minimal experience with genetics laboratory techniques.

D. The focus of their professional education is the study of medicine rather than genetics.

3. How do genetics counselors provide genetic/genomic information to patients and families

in a nondirective manner?

A. Providing only the information the patient or family specifically requests B. Skillfully directing the patient and family toward the best choice that is supported by appropriate research C. Presenting all facts and available options in a manner that neither promotes nor excludes any legally permitted decision or action D. Filtering management options and focusing on the information that will support the decision they believe is right for the individual patient/family 4. A certified family nurse practitioner with an MSN degree in family practice who works in a clinic

serving patients who have connective tissue disorders refers to himself as a clinical geneticist. Is this title appropriate?

A. Yes, he is an advanced-practice nurse.

B. Yes, he is a clinician employed in a setting specializing in patients with genetic-based health problems.

C. No, a clinical geneticist must be certified as a genetic counselor.

D. No, a clinical geneticist is a physician who has completed a fellowship in clinical genetics.

5. Why do genetic counseling programs include extensive courses on laboratory methods in genetics?

A. So that the certified genetics counselor can serve as a backup genetics technician in

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

small laboratories B. So that the certified genetics counselor can perform standard karyotyping on routine blood specimens C. So that the certified genetics counselor can help patients understand testing procedures and results D. So that the certified genetics counselor is able to draw blood proficiently and safely 6. Which genetics professional has the defined responsibility for overseeing the work in a

cytogenetic laboratory?

A. Clinical laboratory geneticist B. Genetic laboratory technician C. Cytogenetics technician D. Medical geneticist 7. A woman’s family history for breast cancer includes two paternal aunts who developed breast

cancer before age 45. Which genetics professional would be most appropriate for assistance in helping this patient understand the health risk posed by this family history?

A. Medical geneticist B. Genetic counselor C. Clinical geneticist D. Clinical laboratory geneticist 8. Which statement or criterion is a required component for certification as an Advanced-

Practice Nurse in Genetics (APNG)?

A. A 2-year residency in clinical genetics B. An earned PhD or DNP with coursework focusing on genetics C. Completion of 500 hours of direct bedside care for patients with genetic disorders D. Completion of a 50-case log describing the nurse’s actions that reflect the standards of clinical genetic nursing practice 9. The patient who has been found to have a mutation in a gene allele that greatly increases her risk

for a serious health problem has asked a generalist nurse to be present when she discloses this information to her family. What is the nurse’s role in this situation?

A. Primary health-care provider B. Genetic counselor C. Patient advocate D. Patient support 10. Which activity would a general registered nurse be expected to perform as part of genomic care?

A. Calculating recurrence risk for parents who have just had a child

with nondisjunction Down syndrome B. Informing a patient that his test results are positive for a genetic disorder C. Obtaining an accurate family history and physical assessment data D. Requesting a consultation visit from a clinical geneticist

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

11. Jessica (aged 32) is BRCA1 positive, which is known to greatly increase the risk for breast and

ovarian cancer. She was tested because her mother, who had ovarian cancer, was BRCA1 positive. Jessica has decided to have both of her ovaries removed because she believes that, in her family, being BRCA1 positive increases the risk for ovarian cancer only. How should a genetic counselor respond to Jessica’s statement?

A. Encourage her to consider a bilateral mastectomy.

B. Accept Jessica’s explanation as a manifestation of her autonomy and remain nondirective in the interactions with her.

C. Clarify that a BRCA1 mutation does not preferentially express ovarian cancer over breast cancer in any given family.

D. Discuss Jessica’s responsibility to inform all the other female members of her family about her BRCA1 status and cancer risk.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 18: Health Professionals and Genomic Care Answer Section MULTIPLE CHOICE 1. ANS: B

Genomic care is ensuring that the influence of a person’s genetic history on health and disease is considered as part of general assessment information for all patients and families. This does not mean that all patients should have some sort of genetic testing. Rather, it means that all healthcare professionals are obligated to avoid overlooking genetic issues that may affect an individual’s health or risk for health problems. The patient determines with whom, if anyone, to share genetic test results. Calculation of odds ratios and recurrence risks is not part of general health care.

PTS: 1 2. ANS: D

The title of genetics professional implies that the individual has extensive education and often, special credentialing in some aspect of the broad genetics field. By these criteria, a person with an entry-level degree in a health-care profession, such as registered nurse, registered dietitian, physical therapist, pharmacist, or physician/surgeon, is not a genetics professional because genetics was not the focus of his or her professional education. Although knowledge of genetics laboratory techniques is helpful to genetics professionals, clinical geneticists are not expert technicians. Acute health problems often have a genetic input to the disorder, as does the patient’s response to therapy.

PTS: 1 3. ANS: C

Nondirective counseling provides all the relevant facts and options available and allows the patient/family to make the decision that is right for them. The counselor does not recommend any course of action but supports whatever action the patient and/or family decides to take. This type of counseling almost always provides more information than the patient requests because most patients have little background in genetics.

PTS: 1 4. ANS: D

Only a physician who has completed training in a clinical specialty residency, such as pediatrics, internal medicine, obstetrics/gynecology, or another relevant specialty, and completed a clinical genetics fellowship in a program accredited by the American Board of Medical Genetics can be called a clinical geneticist. Additionally, the clinical geneticist must acquire initial certification and continue to remain current within the specialty to maintain certification. An MSN-prepared nurse practitioner, even with a certified specialty in genetics, is not a clinical geneticist.

PTS: 1 5. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Genetic counseling programs do include extensive laboratory methods courses, not because a certified genetic counselor (CGC) is expected to perform these tests as part of his or her role, but to ensure that the counselor has adequate background to help patients and families understand testing procedures and results. This is especially important for those tests that require weeks to complete. Genetic counselors do not draw blood, nor do they perform any laboratory techniques, including generating a karyotype, unless they have also completed additional training and are officially certified for such actions.

PTS: 1 6. ANS: A

A clinical laboratory geneticist is either a physician with an MD or a DO (doctor of osteopathy) degree or a PhD degree in genetics or relevant biological science. Specialty training for certification is an additional 24-month fellowship in a program approved by the American Board of Medical Genetics (ABMG). These individuals can then be certified by examination through the ABMG in at least one of three subspecialties: cytogenetics, molecular genetics, or biochemical genetics. A technician (laboratory or cytogenetic) performs the actual test under the direction of a clinical laboratory geneticist but does not interpret the tests or warrant the quality of the work. A medical geneticist does not perform or oversee genetic testing.

PTS: 1 7. ANS: B

A genetic counselor focuses on direct communication with and counseling of patients and families at potential risk for genetic problems. Even though a clinical geneticist may also provide genetic counseling, his or her primary responsibilities are diagnosing and clinically managing patients with a wide variety of genetic disorders. A medical geneticist has a doctorate (PhD), most commonly in population genetics or epidemiology. These individuals often work along with genetic counselors to provide accurate recurrence risk information for affected families but have little, if any, preparation in counseling. A clinical laboratory geneticist also has minimal, if any, counseling preparation or experience.

PTS: 1 8. ANS: D

One of the minimum criteria is the completion of a log of 50 cases within 5 years of the application and four written case studies that reflect the standards of clinical genetics nursing practice developed by the International Society of Nurses in Genetics (ISONG). The candidate must also complete 300 hours of supervised genetic practicum experiences as a clinical genetic nurse, with the genetic practice component being greater than 50%. A master’s degree in nursing from an accredited program is required, but a DNP or PhD and a 2-year residency are not.

PTS: 1 9. ANS: D

The nurse is supporting the client emotionally while she tells the family the information she learned about the test results. The nurse is neither interpreting the results nor counseling the patient and family about what steps to take next. This situation does not require an advocate role.

PTS: 1 10. ANS: C

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A genetics professional has extensive education and, often, special credentialing in some aspect of the broad genetics field. Such a professional is an expert in one or more areas of genetics. By this criterion, a person with an entry-level degree in a health-care profession, such as a registered nurse, is not a genetics professional because genetics was not the focus of her or his professional education. However, all health-care professionals are expected to have at least a basic understanding of the general patterns of inheritance and genetic terminology as well as to be able to construct an accurate three-generation pedigree from assessment information.

PTS: 1 11. ANS: C

Clarifying misconceptions is a responsibility that can be met while still being nondirective.

Jessica can still choose to have only her ovaries removed, but she does need to know her risk for breast cancer in order to come to an informed decision. Encouraging her to have a mastectomy is directive rather than nondirective. Only Jessica can determine whom in her family to inform.

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 19: Financial, Ethical, Legal, and Social Considerations Multiple Choice Identify the choice that best completes the statement or answers the question.

1. Preimplantation genetic diagnosis provides parents with which options?

A. The ability to screen normally fertilized embryos for genetic traits after the

first trimester B. The ability to select embryos for implantation that test negative for a

familial disease mutation C. The opportunity to determine how many children they will conceive D. The ability to guarantee that they will have a healthy baby 2. Your patient, Maggie, insists that her mother have genetic testing to determine if her breast

cancer is connected to a mutation in the BRCA1/2 genes. Maggie is concerned about her own risk for getting breast cancer and that of her children, but Maggie’s mom does not want to be tested. What ethical principle is one of several that must be considered in evaluating this case?

A. Maggie’s “autonomy” B. Maggie’s mother’s “right to know” C. The nurse’s “duty to warn” D. Maggie’s mother’s “right to privacy” 3. A patient asks you whether the Genetic Information and Nondiscrimination Act (GINA) means

that his insurance company is required to pay for his genetic testing if he elects to have it done.

What is your best response?

A. “Yes, if you agree to share the test results with your family, your healthcare provider, and your insurance company.”

B. “Yes, if other family members have already been found positive for a disease-causing mutation.”

C. “No, unless testing finds a specific disorder for which a current medical intervention has been proven effective.”

D. “No, it only protects against discrimination and does not require insurance companies to pay for testing.”

4. You are caring for a college professor who has been offered testing for her family’s mutation in

BRCA1. She expresses fear of genetic discrimination as a reason for refusing genetic testing.

What do you tell her?

A. “There is no need to be concerned about genetic discrimination.”

B. “I appreciate your concern, but there is no way your insurance company or employer will ever be able to get your genetic testing results.”

C. “There is now federal legislation banning genetic discrimination, and in addition, we will do everything we can to keep your results confidential.”

D. “There have been no instances of documented genetic discrimination in insurance or employment. This concern is overblown.”

5. Under which condition(s) would genetic testing for predisposition to an inherited disorder in a

minor child be considered reasonable?

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

A. When the family pedigree indicates an autosomal-dominant pattern of inheritance B. When the risk is high and prophylaxis to reduce the severity of the disorder

is available C. When penetrance is high and the expected onset is middle adulthood D. When the mutation within a family is known and is specific 6. A patient whose mother has Huntington disease is considering genetic testing but is not sure

whether she really wants to know if she has the mutation. She asks you what you would do if your mother had the disease. What is your best response?

A. “I would have the test so that I could decide whether to have children or to use adoption.”

B. “I can only tell you the benefits and the risks of testing; you must make this decision yourself.”

C. “Because there is no cure for this disease and testing would not be beneficial, I would not have the test.”

D. “You need to check with your brothers and sisters to determine whether testing for this disease would be appropriate for you.”

7. A scientist is working to develop a genetic test that will screen embryos so that only those

producing tall children with beautiful features will be implanted. What area of genetic work or studies does this example represent?

A. Eugenics B. Cybernetics C. Cytogenetics D. Genetic imprinting 8. What should be told to the patient who has been found to have a genetic mutation that increases

the risk for colon cancer and says he does not want any of his family to know about this result?

A. “It is required by law that you inform your siblings and your children about this result so that they also can be tested and monitored for colon cancer.”

B. “It is not necessary to tell your siblings because they are adults, but you should tell your children so that they can be tested before they decide to have children of their own.”

C. “It is not required that you tell anyone about this result; however, because your siblings and children may also be at risk for colon cancer, you should think about how this information might help them.”

D. “It is your decision to determine with whom, if anyone, you share this test result; however, if you do not tell any of your family members and they get colon cancer, you would be responsible for their development of the disease.”

9. Sometimes health-care providers with information about family members’ genetic risk are

confronted by conflicting ethical principles. Which principle is least likely to conflict with the health-care provider’s “duty to warn”?

A. Autonomy B. Beneficence C. Right to privacy D. Genetic discrimination

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 19: Financial, Ethical, Legal, and Social Considerations Answer Section MULTIPLE CHOICE 1. ANS: B

Preimplantation genetic diagnosis is a process only done in conjunction with in vitro fertilization because it is performed on early embryonic cells that are in a petri dish and have never been in the mother’s body. A group of embryos generated by in vitro fertilization is removed at the eightcell blastocyst stage. Cells from each embryo can be tested to find gene variants causing singlegene disorders or chromosomal problems, or to determine sex. One or two embryos without the identified familial disease are then selected for implantation. This procedure does not guarantee the infant will be otherwise healthy, only that he or she will not have the specific disorder.

PTS: 1 2. ANS: D

Maggie’s mother has a right to keep her genetic information private even though Maggie would like to know if her mother carries a BRCA1/2 mutation. The nurse’s duty to warn does not play a role here, and Maggie’s mother is not interested in her right to know. Although Maggie wants to make an autonomous decision and have her mother tested, her autonomy does not extend to testing done on her mother.

PTS: 1 3. ANS: D

Title I of GINA makes it illegal for health insurers to use clients’ genetic information to make decisions about their eligibility for insurance, the size of their premiums, or the extent of their coverage. Health insurers also cannot use genetic information as evidence of a preexisting condition, and they cannot require that a client have genetic testing. Title II makes it illegal for employers to use genetic information to make decisions about hiring, promoting, or terminating employees.

Neither provision requires insurers to pay for testing; however, some insurance companies have chosen to do so.

PTS: 1 4. ANS: C

This person’s fears are real and probably based on her knowledge of someone having been discriminated against for an identified genetic problem. With the passage of Genetic Information and Nondiscrimination Act (GINA) by the U.S. federal government, this type of discrimination is illegal. Title I of GINA makes it illegal for health insurers to use clients’ genetic information to make decisions about their eligibility for insurance, the size of their premiums, or the extent of their coverage. Health insurers also cannot use genetic information as evidence of a preexisting condition, and they cannot require that a client have genetic testing. Title II makes it illegal for employers to use genetic information to make decisions about hiring, promoting, or terminating employees.

PTS: 1 5. ANS: B

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Predisposition testing of children is not routinely performed unless there is beneficial treatment available. Just having the parents “want to know” is not enough. In most instances, the recommendations are for the child to make his or her own decision regarding genetic testing as an adult. However, when the risk for the disorder is high and beneficial treatment is available, testing may be done in childhood to begin treatment for prevention of the disorder or for severity reduction. An example is the autosomal-dominant disorder of familial adenomatous polyposis (FAP). People who have the genetic mutation develop thousands of colon polyps and have a 90% or greater risk of having early-onset colon cancer. If a child tests positive for this genetic mutation, he or she has yearly screening for colon cancer. After adolescent growth is complete, the recommendation is that the child have a total colon resection to greatly reduce the risk for colon cancer. If the number of polyps is so high even in early childhood that cancer surveillance is too difficult, the colon is removed in childhood.

PTS: 1 6. ANS: B

Any level of genetic counseling requires the counselor to be “nondirective.” The counselor must ensure that the client has adequate and accurate information upon which to base the decision but cannot suggest or direct the client to test or not to test. The client may wish to discuss the issue with his or her family, but ultimately, the decision about testing can only be made by the client.

PTS: 1 7. ANS: A

Eugenics has been defined as working to improve humankind by selectively breeding people who have genes that society would consider “good” and not allowing reproduction of people with genes that society would consider “bad.” If you are selecting only embryos that will produce beautiful children, you are striving for the same purpose. Cybernetics is concerned with the merger of persons and machine. Cytogenetics is the study of chromosomes. Genetic imprinting refers to the chemical silencing of gene expression.

PTS: 1 8. ANS: C

This situation represents an ethical dilemma. The patient does have the right to disclose or not disclose the information discovered through genetic testing. However, in this instance, disclosure could allow other people to check their health status and take preventive actions against the development of cancer. The patient cannot be forced to disclose the results but should be encouraged to consider disclosure. The response described in C keeps the issue open for discussion later. There is no national or final legal point of view on the resolution of this controversy.

PTS: 1 9. ANS: D

The ethical principles of “right to privacy,” “autonomy,” and “beneficence” can all conflict with the health-care provider’s duty to warn. If we support a patient’s wishes not to share his or her genetic information with family members, we are respecting the patient’s “right to privacy” and “autonomy.” What is “good” for the patient (respecting the right to privacy) may conflict with what is “good” for the family members (being warned). This creates a problem for respecting beneficence. Genetic discrimination is a social problem but not an ethical principle.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 20: Genetic and Genomic Variation Multiple Choice Identify the choice that best completes the statement or answers the question.

1. What is the main purpose of population genetics?

A. Determining the factors that allow allelic frequencies to change over time B. Determining the geographic origins of specific genetic-based disorders C. Assessing the effects of assortive mating on natural selection and evolution D. Assessing the differences between race and ethnicity for susceptibility

and resistance to infectious diseases 2. What factors could increase genetic diversity in a particular population?

A. Genetic drift B. The population effect C. The bottleneck effect D. Increased number of haplotypes 3. What criteria must a population meet in order to stay in Hardy–Weinberg equilibrium?

A. Random mating, no migration, and no mutation B. Founding commonalities and no haplotype differences C. Assortative mating, migration, and frequent mutation D. Limited procreation, no diet change 4. What pieces of genetic information tend to be passed down from generation to generation with the

least variation?

A. Mitochondrial DNA B. Nuclear DNA C. Ribosomal DNA D. Histone proteins 5. Why is it important to consider population genetics?

A. Natural selection can increase genetic diversity.

B. Accurate assessment of a person’s ethnicity can be identified from DNA. C. Disease risk can vary as a result of the geographic origin of one’s ancestors.

D. Knowing ethnicity allows accurate prediction of Huntington disease risk.

6. Which practice is most likely to result in a change in the Hardy–Weinberg equilibrium

of a population or geographic area?

A. Random mating from within the established population B. Geographic isolation of the established population C. Assimilation of immigrants into the existing population D. Preponderance of autosomal-dominant traits in the existing population 7. A group of eight space travelers, four men and four women, settled on the planet Zebulon. Their

descendants had a very high rate of the autosomal-dominant disorder moonophilia distractens.

What factor could explain this phenomenon?

A. Equal exposure to an environmental mutagen

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

B. Hardy–Weinberg equilibrium C. Variable expressivity D. Founder effect 8. A small group of people left their homeland and set sail for a tropical island. They settled there,

and their descendants lived for many generations. Unfortunately, a relatively high proportion of this new population is afflicted with an autosomal-recessive disease. What would explain this?

A. They encouraged immigration of people from the mainland.

B. An unidentified environmental radiation source was present on the island.

C. One of the original group members had the gene mutation from conception.

D. Their lack of genetic diversity made them more vulnerable to new mutations.

9. The Black Death was a pandemic spreading across Europe between 1348 and 1350. Estimates

state that 30% to 60% of Europe’s population died from the Black Death. If we look at Europe’s population before the pandemic and compare it to the population several generations later, what are we likely to find?

A. More genetic diversity in later generations B. Less genetic diversity in later generations C. Less genetic diversity in earlier generations D. The same degree of genetic diversity in later as in earlier generations 10. Why are people of Ashkenazi Jewish descent more likely to be carriers of the mutations that

cause Tay–Sachs and Gaucher disease?

A. The environment of Eastern Europe increased their risk of developing a mutation.

B. The common diet shared by these people has reduced their genetic diversity.

C. Bottleneck effects have reduced the genetic diversity in this population.

D. Being heterozygous for these diseases allowed them to survive cholera.

11. Which term refers to a random change in allele frequencies, not based on natural selection?

A. Population bottleneck B. Genetic drift C. Founder effect D. Migration effect 12. Which statement regarding genetic diversity is most accurate?

A. Larger genes are more likely to display diversity than small genes.

B. Genetic diversity is significant only when a population is isolated.

C. Genetic disorders are more common in populations that have greater

genetic diversity.

D. Population bottlenecks result in loss of alleles that provide minimal

selection advantage.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Chapter 20: Genetic and Genomic Variation Answer Section MULTIPLE CHOICE 1. ANS: A

The field of population genetics examines the ways in which allele frequencies change in human populations over time, including those events that keep the frequencies the same and those events that change them.

PTS: 1 2. ANS: D

Haplotypes are groups of genes or gene variations that are inherited together as genetic neighborhoods. As the number of these haplotypes within a given population increases, so does the genetic diversity.

PTS: 1 3. ANS: A

The requirements for populations to stay in Hardy–Weinberg equilibrium are as follows: • People mate randomly.

• The population is extremely large.

• Everyone has children.

• There are no mutations.

PTS: 1 4. ANS: A

Mitochondrial DNA and the DNA on the Y chromosomes tend to be passed down from generation to generation with very little variation. Processes such as independent assortment of alleles during meiosis and crossing over of segments of homologous chromosomes result in increased genetic diversity.

PTS: 1 5. ANS: C

Disease risk can vary depending on the geographic origin of one’s ancestors. For example, people of Ashkenazi Jewish descent are more likely to be carriers of a number of autosomal-recessive diseases. We cannot accurately identify a person’s race from his or her DNA, and knowing ethnicity does not accurately predict Huntington disease risk. Natural selection decreases genetic diversity.

PTS: 1 6. ANS: C

The first criterion for Hardy–Weinberg equilibrium is no migration (out from or into the existing population. Thus, the arrival of a group of immigrants who are assimilated into the established population represents a loss of the first criterion. As assimilation occurs, even with random mating, the new population adds to the existing gene pool. Geographic isolation of the established population contributes to Hardy–Weinberg equilibrium by reducing the chances for migration. A preponderance of autosomal-dominant traits neither contributes to nor takes away from equilibrium.

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

PTS: 1 7. ANS: D

Founder effect occurs when a small group of people leave a larger population and settle somewhere else and are the main people who end up populating an area. Hardy–Weinberg equilibrium exists when allele frequencies in a population meeting certain criteria remain the same. Heterozygosity refers to the proportion of a population who are heterozygous at a particular locus. Variable expressivity refers to differences in disease severity among people who have the same genotype at the locus of interest.

PTS: 1 8. ANS: C

Even if only one member of the original group was heterozygous for a specific disease-causing gene allele, children from this limited population were likely to intermarry, increasing the incidence of that particular haplotype. By the fourth generation and beyond, so many of the population would be carriers of this disease-causing allele that its expression increased. Some modern examples of this process include groups who have socially isolated themselves and intermarried to such an extent that recessive disorders appear dominant because of the increased incidence, such as the old-order Amish, Ashkenazi Jews, and French Canadians from the Quebec area.

PTS: 1 9. ANS: B

We would find less genetic diversity in later generations. The Black Death caused a population bottleneck, reducing genetic diversity in subsequent generations.

PTS: 1 10. ANS: C

At different times in history, the Jewish population of the world was decreased deliberately, causing a population bottleneck. The holocaust of World War II is just the most recent example.

With reduced numbers of people within that population, only the traits of the survivors were passed on (fewer haplotypes), limiting the genetic diversity and increasing the likelihood for expression of recessive genes.

PTS: 1 11. ANS: B

Genetic drift is defined as a random change in allele frequencies not based on natural selection.

A population bottleneck occurs when some event severely reduces the number of individuals in a population. The founder effect occurs when there is a reduction in genetic variability that comes from the separation of a population subgroup and the reproduction of that less diverse subgroup.

The migration effect is not a term commonly used in population genetics.

PTS: 1 12. ANS: A

Test Bank for Genetics and Genomics in Nursing and Health Care 2nd Edition by Theresa Beery, Linda

Genetic diversity in the form of single-nucleotide polymorphisms is much more common in alleles of larger genes than those of smaller genes (one sequence variation for every 200 to 500 nucleotides). Lack of genetic diversity occurs in isolated populations. Genetic disorders are more common in populations with less genetic diversity. When diversity is greater, there are more alleles for genes, reducing the likelihood of inheriting two alleles that cause disease or disorders.

When population bottlenecks occur, alleles are lost randomly, without regard to whether the allele confers a selection advantage or not.

PTS: 1