Advanced Problems In Genetics Quiz Answers

QUIZ QUESTIONS AND ANSWERS 1. The human genome contains: A. 100,000 genes B. 75,000 genes C. 52,000 genes D. 20,000 genes E. 6,000 genes The human genome contains between 20,000 and 25,000 individual genes (Beery and Workman, 2012). 2. The core unit of chromatin packaging is called _____ and consists of _______. A. nucleus, DNA B. nucleolus, proteins C. nucleosome, histones D. helix, DNA E. centrosome, microtubules The answer to question #2 is C. nucleosome, histones .Histones are special proteins that come in groups of eight. They are arranged in four stacks of two. They are positively charged. The phosphate groups are negatively charged; therefore, they are attracted. The DNA double helix wraps twice around each group of histones. This forms a nucleosome (Osmosis, n.d.). “The DNA molecule of a chromosome exists in chromatin as a complex with a family of basic chromosomal proteins called histones” (Nussbaum et al., 2016, Structure of Human Chromosomes section). DNA in a cell is 2 meters long if it were to be stretched out. Therefore, the DNA is packed into the nucleus (Jorde & Bamshad, 2020). “Each complex of DNA with core histones is called a nucleosome (see Fig. 2-5 ), which is the basic structural unit of chromatin, and each of the 46 human chromosomes contains several hundred thousand to well over a million nucleosomes” (Nussbaum et al., 2016, Structure of Human Chromosomes section). 3. Formerly known as “junk” DNA, this type of chromatin is highly condensed and likely contains many regulatory elements important for gene transcription. A. Euchromatin B. Heterochromatin C. Nucleosomes D. Histones E. 4. Nucleolus While the eukaryotic genome consists of mitochondrial and nuclear DNA, the mitochondrial genome is unique in that. A. It does not replicate B. It lacks introns C. It contains no coding sequences D. It is inherited through the cytoplasm of the sperm E. 5. It associates with different histones than nuclear DNA DNA has a net negative charge at physiological pH due in large part to: A. the DNA bases B. the deoxyribose sugar C. the phosphodiester backbone D. hydrogen binding E. histones 6. If DNA you isolated consisted of 40% adenine, what would be the percentage of cytosine in this molecule? A. 5% B. 10% C. 20% D. 40% E. 60% 7. Which of the following events is in the correct order for DNA replication? A. Nucleosome disassembly, DNA helicase unwinding of DNA, replication by DNA polymerase delta, nucleosome reassembly B. DNA helicase unwinding of DNA, nucleosome disassembly, replication by DNA polymerase delta, nucleosome reassembly C. replication by DNA polymerase delta, DNA helicase unwinding of DNA, nucleosome reassembly D. replication by DNA polymerase alpha of leading strand, replication of lagging strand by DNA polymerase delta, DNA helicase unwinding of DNA E. topoisomerase activity, DNA ligase, DNA helicase unwinding, DNA replication by DNA polymerase delta The correct answer is A. Nucleosome disassembly, DNA helicase unwinding of DNA, replication of DNA polymerase delta, nucleosome reassembly. DNA replication begins with nucleosome disassembly, which is unpacking of DNA from chromatin. The double stranded DNA is separated by and unwound by the enzyme, DNA helicase. DNA helicase breaks the weak hydrogen bonds (Beery & Workman, 2012). This creates the replication fork which is two separated strands that are the template strands for the new strands. There is a leading strand and a lagging strand. In the leading strand, the enzyme, DNA polymerase adds new complementary nucleotides (A, C, G, and T) to the strand in the 5’ to 3’ direction. DNA ligase seals the strands together at the end of replication (Your Genome, 2016). 8. DNA replication is _______ and involves the formation of a bond between the ____ of one nucleotide and the _____ of another nucleotide. A. Uni-directional, 3’ hydroxyl group of the sugar, 3’ phosphate B. Uni-directional, 5’ hydroxyl group of the sugar, 3’ phosphate C. Bi-directional, 3’ hydroxyl group of the sugar, 5’ phosphate D. Bi-directional, 5’ hydroxyl group of the sugar, 3’ phosphate E. Bi-directional, 3’ hydroxyl group of the sugar, 3’ phosphate The answer must be C. Bi-directional, 3’ hydroxyl group of the sugar, 5’ phosphate. The rationale for this is as follows- The answer cannot be B or D because only the 3’ end offers the hydroxyl group (-OH), not the 5’ end (Beery and Workman, 2012). The answer cannot be A because DNA replication is not unidirectional. Lastly, the answer cannot be E because the answer has 3’ phosphate and 3’ hydroxyl. One side is 5’ and one side is 3’. The numbers 3 and 5 refer to specific carbon on the sugar molecule that connects to the phosphorus molecule (Jorde et al., 2020). DNA replication is “bidirectional” because DNA polymerase, the enzyme that adds nucleotide bases, moves only 3’ to 5’ direction. There are two DNA template strands running in opposite directions that need to be replicated and make up the replication fork. One of the new strands that forms is the leading strand (Lumen Microbiology, n.d.). DNA polymerase replicates large portions of the leading strand much faster than the lagging strand because it is smoothly replicating in the direction of 3’ to 5’. However, the other template strand is running antiparallel so it is slower and called the lagging strand. This strand is replicated in chunks at primer sites. These newly synthesized DNA chunks are called Okazaki fragments and are joined together by an enzyme called DNA ligase (Lumen Microbiology, n.d.).  in the 9. Which of the following answers correctly labels #1, 2 & 3 on the replication fork? A. 5’, 3’, 5’ B. 3’, 3’, 5’ C. 3’, 3’, 3’ D. 5’, 5’, 5’ E. 10. 3’, 5’, 5’ In the replication fork above, what enzyme is responsible for “gluing” together the short DNA fragments in strand #2? A. DNA helicase B. Topoisomerase C. DNA polymerase alpha D. DNA ligase E. RPA 11. You give a patient with a urinary tract infection ciprofloxacin (a fluoroquinolone) which inhibits bacterial DNA gyrase. Which of the following statements explains the bactericidal effects of this drug? A. The bacteria are unable to disrupt the hydrogen bonds required for DNA unwinding prior to replication B. The bacteria are unable to relieve the torsional stress in DNA created by DNA unwinding C. DNA polymerase is unable to attach to DNA D. The bacterial Okazaki fragments are unable to be attached together following replication E. The bacteria are unable to initiate DNA replication 12. Why do telomeres get shorter with each round of DNA replication? A. Because DNA polymerase depletes the available free nucleotides required to complete replication B. Because the telomeres form t-loops (circular structures) that protect them from degradation, but limit their ability to serve as templates for replication C. Because the telomeres are easily degraded by DNA exonucleases D. Because removal of the RNA primer, followed by ligation of the lagging strand results in a gap. 13. Chromatin remodeling: A. Is essential for the access of replication proteins to DNA B. Involves the degradation of histones C. Is irreversible D. Occurs in the mitochondria and nucleus E. Occurs only during M phase of the cell cycle 14. In which stage of the cell cycle, do somatic cells most often reside? A. G0/G1 B. S C. G2 D. M 15. One of the key differences between mitosis and meiosis is: A. Mitosis results in the formation of 4 haploid daughter cells B. Chiasmata form during meiosis D. Meiosis exclusively occurs in somatic cells E. Mitosis exclusively occurs in gametes Molecular Biology 1. Place the following events in proper sequential order: Addition of 5’ Cap to mRNA 2___ Transcriptional termination 1___ Export of mRNA from nucleus to Cytosol 3___ A. B. C. D. 3,2,1 3,1,2 1,2,3 2,1,3 For questions 2-5 match the following terms with the correct description 2. Coding sequence= Exon 3. Region of DNA found upstream of coding sequence that plays important role in initiating expression= Promoter 4. Short consensus sequence the facilitates assembly of transcriptional proteins= TATA Box 5. DNA binding protein that plays important role in initiating transcription= Transcription Factor a. b. c. d. e. f. Promoter Exon Intron Transcription Factor TATA Box RNA polymerase II The answer is b. Exon for 2. Coding sequence. The coding sequence is a link of amino acids. The exons are the remaining part of the sequence that code for proteins. Through gene splicing the introns are spliced out and exons are left (Jorde et al., 2020). This is part of the intron-exon structure and what distinguishes eukaryotes from prokaryotes. The exons are covalently bonded together and form the mature messenger RNA (mRNA) (Beery & Workman, 2012). 6. You measure the expression of mRNA from a coding sequence. You find that there are 100 mRNA molecules generated from this sequence. Based on this information: A. B. C. D. E. There will be 100 proteins that will be translated from these mRNA There will be 10 proteins that will be translated from these mRNA There will be 1 protein that will be translated from these mRNA There will be no protein translated from these mRNA You cannot determine how many proteins will result from these mRNA The answer is E. You cannot determine how many proteins will result from these mRNA. According to Beery and Workman (2012), each mRNA molecule has the ability to translated more than once and “multiple ribosomal complexes attach and begin translation as soon as the first ribosomal complex moves down the mRNA away from the start signal” (Beery and Workman, 2012, p. 32). Beery and Workman (2012) use the example of 100,000 insulin; just because 100,000 molecules of protein insulin are needed that does not mean 100,000 insulin mRNA are needed. Protein synthesis is a process that is efficient (Beery and Workman, 2012). 7. Translation involves the recognition of codons on mRNA by the ______ of _____. A. B. C. D. E. small subunit, ribosome large subunit, ribosome amino acid, tRNA anticodon, tRNA anticodon, ribosome 8. The mechanism of action of tetracyclines is related to their ability to bind to the small subunit of the prokaryotic ribosome and block binding of the amino acyl-tRNA to the ‘A’ site of ribosomes. Because of this, A. B. C. D. E. 9. Bacterial cells cannot incorporate new amino acids onto the growing polypeptide chain These drugs directly inhibit peptide bond formation The completed polypeptide chain cannot be ejected from the ribosome The completed polypeptide chain will not be able to be properly folded The mRNA cannot access the ribosome Protein folding A. Is determined by the formation of peptide bonds between amino acids after ejection from the ribosome B. Is determined by primary amino acid sequence of the polypeptide and non-covalent interactions between amino acids C. Is determine by the orientation of amino acids in the “E” site of the ribosome D. Occurs during the process of peptide bond formation within the ribosome E. Occurs only after the polypeptide enters the proteasome 10. Post-translational modifications of proteins: A. B. C. D. E. Include phosphorylation, methylation and acetylation of amino acids Include peptide bonds and non-covalent interactions generally inactivate proteins occur within the ribosome during translation are irreversible 11. Signal transduction involves: A. B. C. D. E. The binding of a ligand to a membrane-bound receptor The binding of a ligand to a cytosolic receptor The use of second messengers to transduce the signal to the nucleus Cross-talk between different signaling pathways All of the above answers 12. An example of a second messenger molecule used in signal transduction is: A. B. C. D. E. Sodium Phosphate cAMP ATP mRNA For questions 13-15 match the following terms with the correct description a. DNA sequencing- provides the most detailed sequence information b. Real TimePCR- Can be used in the TaqMan assay to detect SNPs in a DNA sample c. RFLP- Employs restriction endonucleases 13. Provides the most detailed sequence information- a. DNA sequencing The answer is a. DNA sequencing. DNA sequencing is the most accurate and specific test that is used to find gene variants (Beery & Workman, 2012). It can tell scientists information about “the nature of a specific mutation, the function of a gene, and the gene’s degree of similarity to other known genes” (Jorde et al., 2020, p. 44). There are two methods of DNA sequencing: the Sanger Method and the Next-Generation DNA Sequencing Method (NGS). Although the Sanger method is slow, it is accuract. NGS is the newer method and provides “the most detailed sequence information of the technologies” invented (Whitt, n.d., Section DNA sequencing). The method involves using PCR, adding nucleotides and dideoxy nucleotides, which can be fluorsecently labeled.The last step involves gel electrophoresis to separate by size (Jorde et al., 2020). 14. Employs restriction endonucleases - c. RFLP 15. Can be used in the TaqMan assay to detect SNPs in a DNA sample- b. Real Time PCR 16. CLIA approves which aspect of genetic tests? Analytical validity a. Clinical validity b. Clinical utility c. Cost validity d. Marketing materials e. The diagram above is a pedigree of three generations showing the occurrence of two genetically transmitted diseases. Phenotypes of affected individuals are shown. Circles represent females, and squares represent males. Assume that the diseases are rare, unlinked, and that the individuals not blood related to I-a and I-b (“in-laws) do not have either disease-producing allele. 1. Which one of the following is true concerning individual I-a? a. He is homozygous for Disease 1 and the allele for this disease is autosomal recessive b. He is heterozygous for Disease 1, and the allele for this disease is X-linked recessive c. He is heterozygous for Disease 1 and the allele for this disease is dominant d. None of the above is necessarily true The answer is c. He is heterozygous for Disease 1 and the allele for this disease is dominant. This can be determined by following clues that help determine the pattern of inheritance. 2. The genotype of individual I-b for Disease 2 is: a. Homozygous recessive b. Heterozygous c. Homozygous dominant d. Not enough information to determine The answer is b. Heterozygous. It can be determined that Disease 2 is recessive because it skips generations (Nussbaum et al., 2016). We can see also from II-c and II-d that they are both unaffected for Disease 2 but have offspring with Disease 2. This is another indication that it is recessive. In Autosomal Recessive diseases heterozygous carriers are more common than affected homozygous (Nussbaum et al., 2016). Autosomal Recessive diseases usually occur from two heterozygous parents, further confirming I-b is heterozygous. Children II-h and II-k are affected with Disease 2. That makes 25% of I-a and I-b’s children affected: another indicator I-b is Heterozygous. If he was homozygous recessive, it is likely more children would be affected by Disease 2 (Berry and Workman, 2012). While analyzing this pedigree, it also must be taken into account that the I-b and I-a’s parents do not have either of the alleles for Disease 1 or 2. 3. If individual III-r marries a normal man with a homozygous wild-type genotype, what is the probability that their son would have Disease 2? a. 0 b. ¼ c. ½ d. ¾ 4. If Individual III-v marries a normal man, what is the probability that their son will have both diseases? 5. a. 0 b. 1/16 c. 1/8 d. ¼ What is the probability that offspring produced by a mating between Individuals III-j and III-m would be affected by either Disease 1 or 2? a. ¼ b. ½ c. ¾ d. 1 The questions below do not related to the genogram. 6. A dominantly inherited trait affects a child and his grandmother, but neither parent. This best illustrates which of the following principles? a. Variable expressivity b. New mutation c. Somatic mosaicism d. Nonpenetrance e. Germline mosaicism The answer is D. Nonpenetrance. When someone has the genotype that results in a phenotype but does not express the phenotype. A common example is BRCA-1, the breast cancer gene. Only 80% of people with this mutation will exhibit breast cancer throughout their life; therefore, it has 80% penetrance (PremedHQ Science Academy, 2016). Penetrance is a proportion of a population. Sometimes the penetrance is related to age as in Huntington’s Disease. Penetrance is not a usual Mendelian law so it can make heredity patterns more complex (PremedHQ Science Academy, 2016). Dominant inheritance does not skip generations. Therefore, in this case neither parents expressed the phenotype at all and this was due to nonpenetrance. They did not show any signs of the disease. Penetrance is not like variable expressivity where the person can show some of the phenotype, at a lesser degree (Nussbaum et al., 2016). Penetrance is a complete lack of expression. Other terms for nonpenetrance are reduced penetrance or incomplete penetrance (Nussbaum et al., 2016). Another common dominant disorder where reduced penetrance is seen is in Retinoblastoma. Retinoblastoma has a penetrance of 90% (Jorde et al., 2020). 7. Two parents with autosomal recessive albinism have a child who does not have albinism. This is best explained by: 8. a. An incorrect diagnosis in one parent b. Misattributed paternity c. Genetic heterogeneity d. Reversion of one of the mutant alleles to wild type e. Mosaicism in one parent Which one of the following statements about a dominant genetic trait subject to genomic imprinting is NOT true? a. In a given family, only males or females will be affected. b. The trait may appear to skip generations. c. Affected individuals will always inherit the mutation from either the mother or father, but not both. d. Severity of the phenotype will be independent of the sex of the affected individual. e. 9. One copy of the associated gene may be methylated. The phenomenon of anticipation is associated with which one of the following types of mutation? a. Frameshift b. Missense c. Imprinting d. Triplet repeat expansion e. Deletion 10. When affected males pass a disorder to their daughters but not their sons, what is the likely pattern of inheritance? a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked You will be assigned to provide the initial answer to one of the questions. You may use any resource that you wish to find the answers to the questions and be sure to post your references. Please use the discussion board to discuss the questions with your classmates and faculty. You should post at least 3 times and take the quiz to receive full credit for the DB assignment. I. Clinical Cytogenetic questions (Content from session 6) ● A neonate is diagnosed as having a sub-microscopic-deletion of chromosome 22q11. Name the laboratory tests that may have been done to identify the micro-deletion. ● Describe the change to the normal human chromosome pattern designated by the following: a) 45, X b) 47, XY, +18 c) 47, XY, +21 ● How would the following be described in a cytogenetic report, using ISCN nomenclature? a) A normal female chromosome arrangement b) A chromosome arrangement indicative of Klinefelter syndrome c) A chromosome arrangement indicative of Patau syndrome ● ● Explain the origin, pathology, and significance of a Philadelphia chromosome. Describe Robertsonian Translocations and explain the significance of this for offspring of persons who have this. ● Describe three ways that triploidy can occur. II. Prenatal Diagnosis & Intervention ● What is a neural tube defect? How can it be detected? ● How are fetal cells in the maternal circulation used for genetic diagnosis? ● Describe amniocentesis? What is it used for? When is it indicated? What are the advantages and disadvantages of this procedure? ● What can maternal serum alpha-fetoprotein levels be used to screen for? ● Describe chorionic villus sampling? What is it used for? What are the advantages and disadvantages of this procedure? ● Describe techniques used in invitro fertilization genetic diagnosis. What are the advantages and disadvantages of these techniques? ● Describe ultrasound? Can it be used to reliably detect for genetic disorders?