Download WGU C785 BIOCHEMISTRY UNIT EXAM 2026 SCRIPT FINAL PAPER CORRECT A+ and more Exams Biochemistry in PDF only on Docsity!
WGU C785 BIOCHEMISTRY UNIT EXAM 2026
SCRIPT FINAL PAPER CORRECT A+
◉ 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
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?
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? Enzymes are specific. Enzymes increase the reaction rate for a reaction. Enzymes are reusable. Enzymes lower the activation energy for a reaction.. 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? Formation of the enzyme-substrate complex Release of the product and enzyme complex Formation of the enzyme-molecule complex Formation of the enzyme-product complex. 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? Competitive Uncompetitive
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? Feedback Competitive Allosteric Uncompetitive. 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
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? The activation energy of enzyme-catalyzed reactions are the same as uncatalyzed reactions
The activation energy of the enzyme-catalyzed reactions only changes in response to temperature. The activation energy of enzyme-catalyzed reactions are lower than the uncatalyzed reaction The activation energy of enzyme-catalyzed reactions are higher than 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? The final enzyme in the pathway The first enzyme in the pathway Any enzyme in the middle of the pathway
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?
◉ 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? PCR can be used to produce the mRNA for the BRCA-1 and BRCA- 2 genes, which can then be sequenced to look for mutations.
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. The PCR process will provide an exponential increase in the amino acid sequence of the gene products. Then the amino acid can be sequenced to look for mutations. PCR primers can be engineered to sequence the BRCA-1 and BRCA- 2 genes and determine whether mutations are present.. 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'
