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WGU C785 Biochemistry OA Questions and Answers 2022

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Which level of protein structure is disrupted through the hydrolysis of peptide bonds? Quaternary Tertiary Primary Secondary - ANSWER Primary The primary structure of a protein is the s... equence of amino acids held together by peptide bonds. Peptide bonds are formed by dehydration reactions and disrupted by hydrolysis. A mutation in the beta-hemoglobin gene, which results in the replacement of the amino acid glutamate in position 6 with the amino acid valine, leads to the development of sickle cell anemia. The structures of glutamate and valine are shown below. If the beta hemoglobin gene in a patient with sickle-cell anemia were to be edited so that the valine in position 6 was replaced with a different amino acid, which replacement for valine would be expected to have the best clinical outcome, in theory, for the patient? (Assume the valine can potentially be replaced with any amino acid other than glutamate.) - ANSWER The original amino acid in a healthy patient is glutamate, which is negatively charged. The mutated amino acid is valine, which is non-polar. Valine is causing sickle cell anemia. The best amino acid to replace valine so that the patient is healthy again would be the one most like glutamate, so any negatively charged amino acid. Secondary, tertiary, and quaternary levels of protein structure can all be impacted by exposing a protein to which treatment? Change of a hydrophobic amino acid to a different hydrophobic amino acid Addition of a reducing agent Placement of the protein in a solution with a low pH Increase in the concentration of the protein in solution - ANSWER Placement of the protein in a solution with a low pH Changes in pH affect hydrogen bonds and ionic bonds. Hydrogen bonds in the backbone of amino acids occur in secondary structure, and both hydrogen bonds and ionic bonds occur in the side chains of amino acids in tertiary structure. An increase in beta-pleated sheet structure in some brain proteins can lead to an increase in amyloid deposit formation, characteristic of some neurodegenerative diseases. What is the primary biochemical process that follows the increase in beta-pleated sheet structure that leads to the development of the amyloid deposits? An increase in glycogen formation in the brain cells Aggregation of the proteins in the brain Secretion of glucagon, leading to excessive ketogenesis An increase in anaerobic metabolism of glucose in the brain - ANSWER Aggregation of the proteins in the brain This question is describing changes in protein structure. Aggregation occurs when proteins clump together inappropriately, causing plaques like amyloid deposits to accumulate. Which level of protein structure is determined by the sequence of amino acids? Secondary structure Quaternary structure Tertiary structure Primary structure - ANSWER Primary structure The primary structure of a protein is simply the sequence of amino acids held together by peptide bonds. Which force is most influential in determining the secondary structure of a protein? Hydrophobic effect Disulfide bonding Hydrogen bonding Electrostatic interactions - ANSWER Hydrogen bonding The secondary structure of a protein is built by hydrogen bonds between the carboxyl groups and amino groups on the backbones of the amino acids. Which amino acid would most likely participate in hydrogen bonds? - ANSWER Amino Acid structure 4 This is a polar, uncharged amino acid due to the OH group on the side chain. Polar, uncharged amino acids containing oxygen or NH groups make hydrogen bonds. Which portion of the amino acid is inside the box? The box is surrounding the section below the Alpha Carbon - ANSWER Side Chain The side chain is the variable group of the amino acid, also called the R group. Every amino acid has the same amino group, carboxylic acid group, and an alpha carbon, but the side chain is different. Which pair of amino acids will most likely interact through hydrophobic forces between their side chains? - ANSWER Both of these amino acids are non-polar and therefore can interact together with a hydrophobic interaction. Please note that the "S" in the amino acid on the right is non-polar, while the "SH" group in answer choice D is polar. The S must have an H to be polar and is otherwise non-polar. Which portion of the amino acid is inside the box? The box is over the Carbon at the Center of the chain - ANSWER Alpha Carbon The alpha carbon is the central carbon on an amino acid that holds together the other groups of the amino acid. It is always attached to the amino group, the carboxyl group, the side chain, and a single hydrogen. It is part of the backbone of the amino acid and is found in every amino acid. Given the following amino acid structure, what is the strongest intermolecular force it would participate in to stabilize a protein structure? Ionic bond Disulfide bond Hydrogen bond Hydrophobic interaction - ANSWER Hydrophobic interaction The amino acid pictured only has CH groups in its side chain, and therefore is non-polar. Non-polar amino acids make hydrophobic interactions. Which change would most likely result in a permanent modification of an expressed protein's function? An increase in the pH of a solution in which a protein is dissolved from 6.5 to 8.0, when it is known that the protein has an optimal activity of pH 7.8 A mutation of the gene for a protein that leads to the substitution of a hydrophobic amino acid with a nonpolar amino acid A mutation of the gene for a protein that leads to the substitution of a nonpolar amino acid with a charged amino acid The mutation of a gene for an enzyme involved in protein synthesis following exposure to X-rays, causing the protein not to be synthesized - ANSWER A mutation of the gene for a protein that leads to the substitution of a nonpolar amino acid with a charged amino acid. The mutation of nonpolar amino acid to a charged amino acid will disrupt the original hydrophobic interaction, permanently changing the function of the protein. Which property of enzymes is illustrated in the final step of the enzymatic cycle? - ANSWER Enzymes are reusable. In the final step of the enzymatic cycle, the product is released and the enzyme is able to bind to a new substrate and begin the cycle again. In the enzyme cycle, which step immediately follows induced fit? - ANSWER Formation of the enzyme-product complex The induced fit refers to the conformational change that the enzyme undergoes when it binds to the substrate to form the enzyme-substrate complex. Therefore, the enzymatic cycle step that occurs after the induced fit is the formation of the enzyme-product complex. Which type of inhibition occurs when a particular drug binds to the active site of an enzyme? - ANSWER Competitive Competitive inhibitors compete with the substrate to bind to the active site of the enzyme. Salivary amylase, an enzyme responsible for partial digestion of carbohydrates, has optimum activity at a pH value of 6.8. What is the impact on the activity if the pH is decreased to 4.0? Significantly increase Significantly decrease Slightly decrease Slightly increase - ANSWER Significantly decrease A drop in pH from 6.8 to 4.0 is a significant change in pH. Recall that the hydrogen bonds and ionic bonds that hold protein structures together can be broken by changes in pH. The disruption in protein structure due to this pH change will also significantly decrease amylase activity. Low levels of glutathione are associated with certain types of ovarian and breast cancers. In the synthesis of glutathione, glutathione accumulates in the cell, binding to an enzyme in the pathway and temporarily preventing the synthesis of glutathione. Which type of inhibition is described by this scenario? - ANSWER Feedback The keywords here are that glutathione accumulates and binds to an enzyme in the pathway to prevent synthesis. Feedback inhibition occurs when a product of a pathway turns into an inhibitor of an enzyme earlier in the pathway. Lipase is an enzyme with an optimum temperature of 98.6°F and an optimum pH of 7.0 in the duodenum in the human body. If a person is experiencing a fever of 99.8°F, what will increase the activity of the lipase enzyme? Decrease in temperature Increase in temperature Decrease of the substrate of the enzyme Increase of pH of duodenum to 8.0 - ANSWER Decrease in temperature An enzyme will have the highest activity when it is under optimal conditions. In this case, the fever of 99.8 is above the optimal temperature, so lowering the temperature will increase activity. The enzyme glucokinase only binds its substrate glucose and converts glucose into the product glucose-6-phosphate. Which property of enzymes is described by this scenario? Specificity Activation energy Induced fit Reaction rate - ANSWER Specificity Enzymes have a high degree of specificity. They will bind to one specific class of molecules and usually catalyze only one type of reaction. Low levels of glutathione are associated with neurological, immunological, and cardiovascular impairments. Two enzymatic reactions are involved in the synthesis of glutathione. In the second enzymatic reaction, glutathione synthetase converts glycine to glutathione. What would potentially decrease risks associated with low levels of glutathione? - ANSWER Increasing the amount of glycine in the diet One way to increase glutathione levels is to increase the activity of glutathione synthetase. Glutathione synthetase activity can be increased by increasing the amount of substrate, or glycine available. Which class of enzymes impacts protein function by temporarily removing a phosphate? Kinase Phosphatase Polypeptide Lactase - ANSWER Phosphatase Phosphatases are enzymes that remove phosphate groups from the substrates. How does the activation energy of enzyme-catalyzed reactions compare to those of corresponding uncatalyzed reactions? - ANSWER The activation energy of enzyme-catalyzed reactions are lower than the uncatalyzed reaction The activation energy is the amount of energy needed to get the reaction started and over the energy hill to form products. Enzymes reduce the amount of energy needed to start the reaction or lower the amount of energy needed to climb the energy hill. A final product of a four-step metabolic pathway serves as a noncompetitive inhibitor, binding to an enzyme in this pathway and temporarily turning off the pathway. Which enzyme is most likely to be targeted by the inhibitor? - ANSWER The first enzyme in the pathway The final product's structure is more unlike the substrate for the first enzyme in the pathway than any other, and thus the least likely to fit the active site of enzyme 1. Which of the following could be a potential ramification of a defective nucleotide excision repair (NER) pathway? Thymine dimers would be fixed by DNA polymerase, increasing the probability of developing skin cancer. Thymine dimers would accumulate and increase the probability of developing skin cancer. Thymine dimers, caused by UV light, would be repaired, decreasing the probability of developing skin cancer. Thymine dimers would persist, decreasing the probability of developing skin cancer. - ANSWER Thymine dimers would accumulate and increase the probability of developing skin cancer. A defective nucleotide excision repair pathway would not be able to repair thymine dimers. The accumulation of thymine dimers would increase the probability of developing skin cancer. A patient has received a large dose of ionizing radiation at his place of employment. Which scenario is accurate? Ligase removes a single damaged base and replaces it with a new nucleotide. Double-stranded DNA breaks are being repaired by nonhomologous end joining, without the use of a homologous template. The homologous recombination pathway is upregulated and is facilitating the removal of mismatched bases. A string of damaged nucleotides is removed and replaced by a new nucleotide sequence in the base excision repair pathway. - ANSWER Double-stranded DNA breaks are being repaired by nonhomologous end joining, without the use of a homologous template. Due to the large dose of ionizing radiation, the patient has many double-stranded DNA breaks, and undamaged homologous DNA is in short supply. Nonhomologous end-joining will be used to fix these double-strand breaks. Immediately following transcription, mRNA must be processed before it is transported to the cytoplasm to undergo translation. Which statement correctly describes mRNA processing? Alternative combinations of introns can be linked together to produce closely related gene products. Exons are removed and introns are inserted into the mature mRNA sequence. mRNA is folded into beta-pleated sheets and alpha helices to produce a mature mRNA sequence. Introns are spliced out and exons are connected to produce a mature mRNA sequence. - ANSWER Introns are spliced out and exons are connected to produce a mature mRNA sequence. Introns are removed from the mRNA sequence and the remaining exons are spliced together to create the mature mRNA. A point mutation has altered the amino acid sequence of a neuronal tau protein, causing serine (Ser) at position 202 to be mutated to proline (Pro). Which set of codons below corresponds to this mutation? TCT to CCT AGA to GGA CGA to AGG AGG to CCC - ANSWER TCT to CCT TCT is a codon for Ser, and CCT is a codon for Pro. TCT to CCT is a point mutation that changes Ser to Pro. Which of the following is required for DNA polymerase to replicate template DNA in PCR? - ANSWER DNA Primers DNA primers are used in the PCR process. A portion of the HLA gene and its associated amino acid sequence are shown below. If cytosine is deleted from codon 101, what will be the resulting amino acid sequence? Ser Tyr Glu Glu Met Leu Glu Pro Cys Glu His Ala Asp - ANSWER Glu Met Leu Deletion of cytosine (C) from codon 101 leads to the DNA sequence 5'-GAG ATG CTG AT-3', which corresponds to the amino acid sequence Glu Met Leu. Inheriting mutations in the BRCA-1 or BRCA-2 gene can increase an individual's likelihood of developing breast cancer or ovarian cancer. How can PCR be used to assess an individual's susceptibility to developing these cancers? - ANSWER PCR primers can be engineered to flank the mutation of interest in the BRCA-1 and BRCA-2 genes, and the PCR product can be sequenced to look for mutations. By using primers that flank the mutation of interest, PCR can produce DNA which can be used to determine if the mutation is present. Which template DNA sequence would result in the amino acid proline (Pro)? 5' - TGA - 3' 5' - TGG - 3' 5' - GGT - 3' 5' - CCC - 3' - ANSWER 5' - TGG - 3' The correct answer is 5'-TGG-3'. The corresponding mRNA for 5'-TGG-3' is 3'-ACC-5', which reversed is 5'-CCA-3', a codon for the amino acid Pro. Legumain is an enzyme that is overexpressed in neurons in response to traumatic brain injury (TBI). How does TBI trigger an increase in the expression of the gene that codes for legumain? - ANSWER Nucleosomes are more widely spaced, exposing the legumain gene. Increased nucleosome spacing leads to increased transcription and translation (i.e. gene expression) of the Legumain gene. Which of the following mRNA sequences can be translated into a peptide sequence containing two amino acids? 5' - GAU CGA UAG UGG - 3' 5' - GAU CGA UGG UAG - 3' 5' - UGG UAC CAG UGA - 3' 5' - GAC CAU GCG GGG - 3' - ANSWER 5' - GAU CGA UAG UGG - 3' The sequence 5' - GAU CGA UAG UGG - 3' contains a stop codon in the 3rd position, therefore, this sequence would produce a peptide sequence with two amino acids. What is the mRNA sequence that would result from this coding DNA sequence: 5' - CAG TTA GAT TCA - 3'? 5' - ACU UAG AUU GAC - 3' 5' - CUG TTU GUT TCU - 3' 5' - CAG UUA GAU UCA - 3' 5' - UGA AUC UAA CUG - 3' - ANSWER 5' - CAG UUA GAU UCA - 3' The coding DNA and mRNA sequences in transcription will be identical, except for any T in the coding DNA sequence having a corresponding to U in the mRNA sequence. The correct answer is 5'-CAG UUA GAU UCA-3'. What is happening to the pH of the lungs due to exhalation? CO2 decreases from exhalation, thus the pH of the lungs increases. [Show More]

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