Biochemistry C 785 Readiness Check 2020 Biochemistry C785 Readiness Check - A Grade

Biochemistry C785 Readiness Check 2020 – Western Governors University Biochemistry Readiness Check The 2nd OA Readiness Check is designed to give you additional practice with new assessment-style questions to help prepare for success on the OA. Thus, please try to mimic exam-like conditions by taking the test in one 2 hr sitting without any study aids apart from your whiteboard/scrap paper. Note: Responses cannot be edited. You will be able to see the results and feedback after submitting the test. A score of 70% is required for OA approval. If you score below 70%, please speak with a Course Instructor via appointment to discuss next steps prior to retaking this test. Points: 54/67 1 What is the complementary sequence to 3’ ATG CGA ATA 5’? (1/1 Points) 3’ TAT TCG CAT 5’ The correct answer is 3' TAT TCG CAT 5'. Remember complementary means “Matching or Pairing” You have to remember to pay attention to your numbers as well as your letters (A-T, G-C, 5'-3'). The correct answer is 3’ ATG CGA ATA 5’ (original sequence) 5’ TAC GCT TAT 3’ (complementary sequence) But we asked for it in the 3’ direction, so mirror answer to give correct answer: 3’ TAT TCG CAT 5’ 3’ TAC GCT TAT 5’ 3’ GAT AGC ATA 5’ 5’ ATA AGC GTA 3’2 Which amino acid sequence would result from this Coding DNA strand? 3’ ATG CAG ATA 5’? (0/1 Points) Tyr Val Tyr The correct answer is Ile Asp Val. We are starting at the coding strand, and have to remember the relationship between coding DNA and mRNA. These two strands are non-complementary and parallel. So we copy the coding strand , change T --- U, and then write the mRNA sequence: 3’ ATG CAG ATA 5’ coding 3’ AUG CAG AUA 5’ mRNA Mirror by changing orientation: 5’ AUA GAC GUA 3’ Read chart Ile Asp Val (chart is in direction of 5' --- 3') If you chose Tyr Val Tyr, this answer is incorrect because this is the amino acid sequence that results from the mRNA 5' UAC GUC UAU 3' which would have been complementary to the given coding strand. Coding DNA is non-complementary and parallel to mRNA. Ile Gln Ile Ile Asp Val Leu His Lys 3 Which amino acid sequence would be made from this template DNA strand? 5’ TAT TAC CGA 3’? (1/1 Points) Tyr Tyr Arg Ile Met Ala Ser Val Ile The correct answer is Ser Val Ile because 5’ TAT TAC CGA 3’ template is complementary and antiparallel so 3’ AUA AUG GCU 5’ but it is in the wrong orientation, so mirror 5’ UCG GUA AUA 3’ and read the chart Ser Val Ile Ser His Gln 4Which of the following are the correct components for a PCR reaction? (1/1 Points) dNTPs, Primer, RNA Polymerase, template RNA dNTPs, Primer, DNA Polymerase, template DNA The correct answer is dNTPs, Primer, DNA Polymerase, template DNA. Notice that all components are about DNA. The Primer is even a DNA Primer. ATP, Primer, mRNA polymerase; template mRNA Acetyl CoA, RNA primer, DNA Ligase, Template phosphate 5 Assuming 100% reaction efficiency, how many DNA copies will be produced after 5 PCR cycles, if we begin with 1 DNA template? (1/1 Points) 16 32 The correct answer is 32. 2x2x2x2x2=32 64 10 6 Which of the following would represent a silent mutation if this is the original sequence: 5’ AUC GUA ACA 3’? (1/1 Points) 5’ AUC GGA ACA 3’ 5' AUA GUA ACA 3' The correct answer is 5' AUC GUA ACA 3' Ile Val Thr 5’ AUG GUA ACA 3’ 5’ AUC GCA ACA 3’ 7If the original coding sequence is 5’ CGA TAC TTC AGA 3’ and it is mutated to 5' CGA TAT TTC AGA 3', what type of mutation would have taken place? (0/1 Points) Silent Missense The correct answer is silent mutation. The nucleotide sequence changes, but it codes for the same amino acid. The coding sequence 5' TAC 3' corresponds to the mRNA sequence 5' UAC 3' (Tyr), and the coding sequence 5' TAT 3' corresponds to the mRNA sequence 5' UAU 3' (Tyr). Since the C changed to at T, this is a point mutation. If the point mutation results in the same amino acid in the new sequence as in the original sequence, the point mutation is a silent mutation. Nonsense Insertion 8 This learning objective is now tested in a different WGU course than Biochemistry. Please select True. True The correct answer is Option 1 because an autosomal dominant disorder would be inherited on numbered chromosomes, not sex chromosomes X or Y. Also, at least one dominant allele (yellow box) needs to be present for the individual to have the dominant disease. False 9 This learning objective is now tested in a different WGU course than Biochemistry. Please select True. True The correct answer is X- linked recessive because parents (carriers) do not have it (II-5- 6) but a child does (III-5). You will get the same result if you consider parents (carriers) (I-1-2), who do not have the trait, but a child does (II-3). A third option that gives thesame result (X-linked recessive) is by considering parents (carriers) who do not have the trait (III-1-2), and their child does (IV-1). The pattern is recessive because the selected parents are carriers, and it is X-linked because only males have the trait. False 10 This learning objective is now tested in a different WGU course than Biochemistry. Please select True. True The correct answer is 50%. Homozygous recessive: aa Heterozygous: Aa When you see "percentage" or "probability," think Punnett square. 50% of the children would be expected to be Aa, and 50% of the children would be expected to be aa. False 11 Which of the following describes an epigenetic change? (0/1 Points) Denaturation of template DNA to facilitate primer annealing. Increased methylation of the promoter region of a tumor suppressor gene in a developing fetus. Thymine dimer formation resulting from UV radiation. The correct answer is "Increased methylation of the promoter region of a tumor suppressor gene in a developing fetus." Mismatch mutation caused by mistakes made by DNA Polymerase during replication. 12 Rett syndrome is a brain disorder that occurs almost exclusively in females, causing severe deficits in language, learning, coordination and other brain functions. Decreased expression of the MECP2 gene causes Rett syndrome. Which of the following scenarios correctly describes how Rett syndrome could be developed? (0/1 Points)A DNA-binding protein blocks RNA Polymerase from binding to the promoter sequence, facilitating the transcription of the MECP2 gene. The answer is "Transcription factors are unable to bind to the transcription start site of the MECP2 gene because nucleosomes are tightly packed together." Think "increased space gives increased access and increased expression." Gene expression is increased when nucleosomes are widely spaced and transcription factors and RNA Polymerase are able to bind to the transcription start site of the gene. In this question, decreased expression is resulting from decreased space between the nucleosomes, so the RNA Polymerase and transcription factors have decreased access to the transcription start site of the gene. If you answered, "A DNA-binding protein blocks RNA Polymerase from binding to the promoter sequence, facilitating the transcription of the MECP2 gene," this answer is incorrect. Transcription factors are proteins that bind to the promoter region on the 5' side of the gene to be expressed. The RNA Polymerase then binds to the transcription start site. Transcription factors are unable to bind to the transcription start site of the MECP2 gene because nucleosomes are tightly packed together. Transcription activators cause nucleosomes to separate, exposing the MECP2 gene. RNA Polymerase binds to the MECP2 gene and begins translation. 13 What happens when the incorrect base is added during the synthesis of a DNA strand in DNA replication? (1/1 Points) The homologous chromosome is used to replace the incorrectly added base with the correct one. DNA Polymerase removes the incorrect base and adds in the correct base. The correct answer is "DNA Polymerase removes the incorrect base and adds in the correct base." DNA Polymerase repairs mismatch errors that occur during DNA replication. Thymine dimers occur.Distortion of the double helix occurs and is repaired by RNA Polymerase. 14 What is the correct definition of nucleotide excision repair? (1/1 Points) Removal of a single damaged nucleotide Damage to a few or several nucleotides are identified, then many nucleotides are removed and all are replaced to repair the DNA segment The correct answer is "Damage to a few or several nucleotides are identified, then many nucleotides are removed and all are replaced to repair the DNA segment." In nucleotide excision repair, several nucleotides are removed whereas, in BER (base excision repair), a single nucleotide is removed. Required when there are breaks in the double stranded DNA strand which causes discontinuity in both strands Insertion of a thymine dimer 15 If arginine is mutated to leucine within a protein, how would the structure of the protein be affected? (1/1 Points) Ionic bonds will continue to form, allowing the protein to fold as normal. Hydrophobic interactions will continue to occur resulting in normal folding. Ionic bonds will no longer form, potentially causing the protein to misfold. The correct answer is "Ionic bonds will no longer form, potentially causing the protein to misfold." Since arginine is a positively-charged amino acid, it would have formed an ionic bond with a negatively-charged amino acid in the protein. Leucine is not charged and is hydrophobic, so it will not form this same ionic bond, and could lead to protein misfolding. Hydrophobic interactions will be broken, potentially causing aggregation. 16Which pair of amino acids below could bond together to stabilize the tertiary structure of a protein? (0/1 Points) Option 1 Option 2 The correct answer is "phenylalanine" and "methionine." Phenylalanine and methionine are both nonpolar (hydrophobic) amino acids, and they interact in the core of the protein through hydrophobic forces. Option 3 Option 4 17 On a summer's day, I was cleaning my garage. My cat Sophie was out with me, and got into a bottle containing a strong reducing agent. Of course I took her straight to the vet. What type of amino acid bonds would be disrupted by exposure to the strong reducing agent? (1/1 Points) Hydrophobic interactions Hydrogen bonds Disulfide bonds The correct answer is covalent (disulfide) bonds. Hydrophobic interactions are disrupted by heat. Hydrogen bonds and ionic bonds are both disrupted by pH and by salt. Ionic bonds 18 What type of bond can the amino acids below form to stabilize the tertiary structure of a protein? (0/1 Points) Ionic bondThe correct answer is "hydrogen bond." Each of the amino acids shown is a polar amino acid, and polar amino acids interact through hydrogen bonds. The NH combined with - - - - - - - - - - - - - - - - - - - - - - - - - - - is 18 weeks pregnant with twins and was recently diagnosed with gestational diabetes. Susie’s twins are developing according to schedule and Susie hopes to carry the twins as close to full term as possible. Which statement below describes how a sufficient amount of oxygen is made available to Susie’s developing fetuses? (1/1 Points) Bicarbonate binds to Susie’s hemoglobin and stabilizes the T-state conformation, which favors the delivery of oxygen to the developing fetuses. 2,3-BPG binds to Susie’s hemoglobin and stabilizes the T-state conformation, which favors the delivery of oxygen to the developing fetuses.The correct answer is, "2,3-BPG binds to Susie’s hemoglobin and stabilizes the T-state conformation, which favors the delivery of oxygen to the developing fetuses." 2,3-BPG binds to hemoglobin and stabilizes the T-state which favors oxygen delivery. Carbon monoxide binds to Susie’s hemoglobin and stabilizes the R-state conformation, which increases the delivery of oxygen to the developing fetuses. Bicarbonate induces Susie’s hemoglobin to adopt a non-planar conformation and bind oxygen so that the developing fetuses are sufficiently oxygenated. 32 Alex is preparing to run in his first 5K. As he stretches, his muscles are producing ATP through aerobic metabolic pathways. A byproduct of a functioning citric acid cycle is carbon dioxide. How is this carbon dioxide removed from the body? (1/1 Points) Carbon dioxide is converted to bicarbonate and binds to hemoglobin, which shuttles the carbon dioxide to the lungs so that it can be exhaled out of the body. Carbon dioxide flows through the bloodstream to the lungs so that it can be exhaled out of the body. Carbonic anhydrase helps carbon dioxide stick to hemoglobin, which shuttles the carbon dioxide to the lungs. In the lungs, Carbonic Anhydrase converts carbon dioxide to a gas so that it can be exhaled out of the body. Carbonic anhydrase converts carbon dioxide to bicarbonate, which can flow through the bloodstream to the lungs. In the lungs, Carbonic Anhydrase converts bicarbonate b-