Test Bank for Genetics and Genomics in Nursing and Health Care, 1st Edition by Theresa A. Beery All Chapters A+

Test Bank for Genetics and Genomics in Nursing and Health Care, 1e by Theresa A. Beery (All Chapters Answer at the end of each Chapter) Test Bank for Genetics and Genomics in Nursing and Health Care, 1e by Theresa A. Beery (All Chapters Answer at the end of each Chapter) 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 correct interpretation of the statement “the HFE gene locus is 6p21”? A. Both alleles of the HFE gene are equally expressed. B. The HFE gene is inherited from the paternal chromosome line. C. The HFE gene alleles are located on the “short arms” of chromosome number 6. D. There is a somatic cell mutation involving gene allele damage on chromosome 6. ____ 4. 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 ____ 5. 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 ____ 6. 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% ____ 7. 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. Autosome ____ 8. What would be the sequence of DNA that is complementary to a DNA section with the base sequence of GGTCAATCCTTAG? A. GATTCCTAACTGG B. TTGACCGAAGGCT C. AACTGGCTTCCGA D. CCAGTTAGGAATC ____ 9. 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 ____ 10. What activity occurs during S phase of the cell cycle? A. The cell undergoes cytokinesis. B. Activity stops and the cell “sleeps.” C. All DNA is completely replicated. D. Chromosomes separate causing nucleokinesis. ____ 11. Which chromosome number represents the euploid state for normal human somatic cells? A. 44 B. 46 C. 47 D. 48 ____ 12. 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. ____ 13. 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. ____ 14. At what phase of the cell cycle are chromosomes visible as separate structures? A. G1 B. G2 C. S D. M ____ 15. Which statement about 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 ____ 16. What is the result of DNA replication? 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 ____ 17. 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. ____ 18. 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. ____ 19. Under what normal condition are genotype and phenotype always the same? A. Trisomy of alleles B. Triploidy of alleles C. Homozygosity of alleles D. Heterozygosity of alleles ____ 20. 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. ____ 21. How does the enzyme DNA ligase 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 ____ 22. 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 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 very 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 function of a specific nucleotide sequence. PTS: 1 3. ANS: C Each gene has a specific chromosome location, called a gene locus. The segments of chromosome extending above the centromere are known as the short arms, or the p arms. The segments of chromosome below the centromere are the long arms, or the q arms. The locus of a gene on a chromosome is pinpointed using the chromosome number, the designation for short or long arm, and number of the specific band region of that short or long arm. PTS: 1 4. 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 5. 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 6. 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 7. ANS: C 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. An autosome is one of the 44 chromosomes that is not a sex chromosome (neither an X nor a Y). PTS: 1 8. 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 9. ANS: B Cytosine and guanine normally pair together, whereas adenine and thymine normally pair together. The reason for this specific and complementary pairings 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, while 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 10. ANS: C The “S” stands for “synthesis.” This is the time during the cell cycle when DNA replicates completely into two identical sets of double-stranded DNA. Cytokinesis is a stage of mitosis. There is no “sleep” stage of the cell cycle, although there is a reproductive resting state (G0) outside of the cell cycle. Nucleokinesis occurs during the M (mitosis) phase of the cell cycle. PTS: 1 11. 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. PTS: 1 12. 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 13. 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 14. 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 15. 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 16. 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 17. 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 18. ANS: C 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 19. 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 20. 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 it. 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 21. ANS: D 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 22. 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. The Punnett Square would show that 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: Mother B Mother O Father A A B A O Father B B B B O PTS: 1 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 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 “sense” strand. ____ 3. 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. ACGCUCGAUUAUUU D. AACGCUUGGUCUGAAUUUUAAUUU ____ 4. What is the function of ribosomes (also known as ribosomal RNA) in protein synthesis? A. Allows interpretation of the two strands of DNA to determine which is the “sense” strand and which is the “antisense” strand B. Serves 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. Allows further processing of synthesized proteins (post-translational modification in order to ensure that the final product is physiologically active) D. Serves as a transport molecule able to move a specific amino acid to the site of protein synthesis (peptide chain elongation) in the correct sequence ____ 5. A strand of recently transcribed messenger RNA contains the following components: exon (1), intron (2), intron (3), exon (4), intron (5). Which sequence represents the mature messenger RNA? A. 1, 4 B. 2, 3, 5 C. 2, 3, 4 D. 1, 2, 3, 4, 5 ____ 6. 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. 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. ____ 7. 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 ____ 8. 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 ____ 9. 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 ____ 10. 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 “sense” strand of DNA is transcribed C. Determining the placement of the “start” signal for translation D. Promoting post-translational modification for conversion to an active protein ____ 11. 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 ____ 12. 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 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. ____ 13. 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 post-translational modifications. D. The introns of one gene may be the exons of another gene. ____ 14. Which DNA segment deletion would cause a frameshift mutation? A. TCT B. GAGTC C. TACTAC D. GCATGACCC ____ 15. 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 ____ 16. 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. ____ 17. What is the expected outcome when a person (twin A) experiences a large deletion of DNA in one of his noncoding region 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. ____ 18. 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. ____ 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. 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. ____ 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. 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 upon which tissue type experienced the somatic mutation. D. Multiple somatic mutations are required for passing on a predisposition to cancer development. 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: 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 actual gene for the protein needed 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 3. 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 4. 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 post-translational modification. The transport molecules are the transfer RNAs (tRNAs), not the ribosomes. PTS: 1 5. ANS: A Converting the early transcript of messenger RNA into mature messenger RNA 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 that is then ready for translation. PTS: 1 6. ANS: C Proteins are not in their final forms for active function when they are first synthesized and require post-translational 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 7. ANS: C 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 8. 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 9. 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 10. 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 11. ANS: D The addition of a poly-A tail to the newly transcripted 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 12. ANS: A 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 13. 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 synthesis of the gene product for which they are introns. PTS: 1 14. ANS: B Because an amino acid is encoded in the DNA by a “triplet” of bases, deletion of any number of bases which 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 15. 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 16. 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 17. 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 18. 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 19. 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 20. 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 Chapter 3: Genetic Influences on Cell Growth, Cell 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. 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. ____ 3. 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 ____ 4. 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 ____ 5. 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 ____ 6. 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 formation of a large nuclear-to-cytoplasmic ratio during the G0 state ____ 7. Normal cells spend most of their life spans in which phase? A. G0 B. G1 C. S D. M ____ 8. 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 ____ 9. 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 ____ 10. Which feature or characteristic of early embryonic cells is unique in comparison with normal differentiated cells? A. Diploidy B. Pluripotency C. Controlled cell division D. Mitosis resulting in four new daughter cells ____ 11. 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 ____ 12. 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 ____ 13. Which stage of cell division is present in mitosis but is missing in meiosis? A. G1 B. S C. G2 D. M ____ 14. 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. ____ 15. 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 ____ 16. 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 ____ 17. 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 ____ 18. 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 ____ 19. 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 oocytes results in a total of eight gametes. ____ 20. How many mature ovum result from the complete oogenesis of one oogonium? A. One B. Two C. Three D. Four ____ 21. 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 ____ 22. 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. Chapter 3: Genetic Influences on Cell Growth, Cell 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: 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 3. 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 4. 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 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. PTS: 1 5. 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 6. 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 7. 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 8. 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 9. 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 10. 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 11. 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 deceases. PTS: 1 12. ANS: A 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 13. ANS: C For meiosis, there is no G2 phase. Cells begin M phase, which is prolonged, immediately after leaving S phase. PTS: 1 14. 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 15. 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 16. 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 17. 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 18. 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 19. 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 20. 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. PTS: 1 21. ANS: D 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 22. 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 Chapter 4: Patterns of Inheritance Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. 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 ____ 2. 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 ____ 3. 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 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. ____ 4. 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%) ____ 5. 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. ____ 6. 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 ____ 7. Which type of genetic transmission promotes the continued existence of genetic mutations in single genes? A. Autosomal dominant B. Autosomal recessive C. Codominant D. Sex-linked ____ 8. 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 higher probability for having a child who does not express the trait at all. B. A person with high expressivity of the trait has a greater risk for 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. ____ 9. What is the most obvious expected finding when analyzing a four-generation pedigree for sex-linked dominant transmission of a trait? A. Transmission is never father to son. B. Females are affected at twice the rate of males. C. An unaffected mother can transmit the trait to her sons. D. An affected mother always transmits the trait to all her children equally. ____ 10. 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. ____ 11. 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 C. Yellow allele dominant; blue allele recessive; sex-linked D. Yellow allele recessive; blue allele dominant; sex-linked ____ 12. 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. ____ 13. 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 disorder to his or her children. D. Because of the possibility of incomplete or low penetrance, an unaffected offspring could have an affected child. ____ 14. 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. ____ 15. 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 expression of identical alleles. C. The mutation occurred in a somatic cell rather than a germ cell. D. Mutation repair is incompletely penetrant. ____ 16. 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 ____ 17. 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. ____ 18. Which genetic disorder has both an autosomal dominant and an autosomal recessive form? A. Syndactyly B. Phenylketonuria C. Long QT syndrome D. Retinitis pigmentosa ____ 19. 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 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. ____ 20. 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. ____ 21. 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. ____ 22. How does histone modification alter gene expression? A. Modified histones result in increased DNA methylation, which increases 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. ____ 23. What is the implication of the Agouti mice experiments with methylation? A. Methylation has the potential to affect the expression of polygenic traits but not monogenic traits. B. Epigenetic changes in gene expression can become more severe with each new generation. C. Some variations in phenotype are not related to DNA allele sequence variation. D. Epigenetic changes are reversible when DNA repair mechanisms are active. ____ 24. Which common health problem has the highest heritability estimate? A. Hypertension B. Peptic ulcer disease C. Congenital heart disease D. Schizophrenic behavior ____ 25. 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 disord