(Test Bank) Genetics and Genomics in Nursing and Health Care

Beery, 2

Nursing

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Test Bank for Genetics and Genomics in Nursing and Health Care, The downloadable file consists of 169 pages, and the total size of the file is 2.26 MB. Purchase now only for highly discounted price.

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Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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? Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company ____ 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. Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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. Copyright © 2018 F. A. Davis Company 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 double-stranded 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 Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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) Copyright © 2018 F. A. Davis Company 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. Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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. D. They will now be dizygotic twins instead of monozygotic twins. Copyright © 2018 F. A. Davis Company ____ 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? Copyright © 2018 F. A. Davis Company A. Increased production of insulin B. Decreased production of insulin C. Increased degradation of insulin D. Decreased degradation of insulin Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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 13. ANS: C Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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. PTS: 1 Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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? Copyright © 2018 F. A. Davis Company 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 Copyright © 2018 F. A. Davis Company 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