BIOCHEM FRQ
4.1 Describe the two ways CO2 has an effect on hemoglobin function.
1. High concentrations of CO2 found near actively respiring tissues, decrease the pH in erythrocytes. This is due to CO2 reacting with H2O to form carbonic acids, which dissociates to increase proton concentration. The resulting lowering of pH induces histidines in globin proteins to be protonated and to become positively charged; this promotes electrostatic interactions (salt bridge formations) between the histidines and negatively charged amino acids(e.g aspartate) This contributes to the stabilization of hemoglobin in the "tense" state lowering its affinity for oxygen's release from hemoglobin 2. CO2 reacts with terminal amino groups of the globin polypeptides to form negatively charged carbamate groups. These carbamate groups stabilize the "tense" form of hemoglobin, deceasing its affinity for binding oxygen and increasing oxygen's release from hemoglobin.
3.1 Name the three amino acids that form the catalytic triad for chymotrypsin and describe the roles each of these amino acids pay in the catalysis of peptide bond hydrolysis.
1.Serine 195: When the polypeptide substrate enters the active site( i.e for cleavage of a peptide bond between a large, hydrophobic amino acids and its neighboring amino acid) serine 195 donates a proton form its side chain hydroxyl to histidine 57 creating an alkoxide ion, making serine a strong nucleophile. Alkoxide ion of serine attacks carboxyl carbon on one side of peptide bond creating an unstable oxyanion hole that induces peptide bond to be cleaved and release of one peptide fragment (with terminal amino group from one side to of cleaved peptide bond) Upon entry of a water molecule, the hydroxyl(created after donation of a proton to histidine) binds with the carboxyl carbon of the remaining peptide fragments, creating a oxyanion hole; the carboxyl carbon bond with serine is broken, releasing the second peptide fragments. Serine accepts a proton back from histidine to re-establish its side chain hydroxyl
4.2 In addition describe how these effects are important for the physiological functions of organism that use hemoglobin for carry oxygen.
3. The negative effects of CO2 has on hemoglobin's affinity for oxygen means that in actively respiring tissues, that utilize and need a lot of oxygen and produce a lot of CO2, hemoglobin will give up a greater proportion of its oxygen than in tissues not producing as much CO2, i.e tissues that aren't as metabolically active. Therefore, these effects of CO2 help ensure that O2 is released from hemoglobin at the highest rate where it is most needed
Address each of the following with respect to a chemical reaction: b. Indicate whether the "activation energy" of a chemical reaction is included in the calculation of ΔG for a reaction; explain your answer.
Activation energy is NOT included in calculation of ΔG because all the energy added to reach transition state is released/recovered when transition state is transformed into the product(s). (ΔG is simply the difference in free energy of products and reactants.)
3.3 Name the three amino acids that form the catalytic triad for chymotrypsin and describe the roles each of these amino acids pay in the catalysis of peptide bond hydrolysis.
Aspartate 102: Forms a hydrogen bond with histidine 57 stabilizing the histidine in the appropriate orientation/position in the enzyme so that histidine can function as a proton acceptor/donor with serine 195 the substrate and water.
Address each of the following with respect to a chemical reaction: d. Describe the effect that an enzyme has on the ΔG of reaction.
D. Enzymes do NOT affect the levels of free energy in the reactants or in the products, and since ΔG is the difference in free energy of products and reactants, enzymes have no effect on ΔG.
Describe, in detail, the structure of DNA. Be sure to include structural elements that make up each nucleotide and the relationships between the nucleotides in a DNA molecule. In addition, indicate how a typical molecule of DNA differs from a typical molecule of RNA.
DNA is composed of four nucleotides: adenosine, guanosine, cytidine, and thymidine Each nucleotide is composed of a base: (adenine, guanine, cytosine or thymine) bound via a beta-glycosidic linkage to the 1' carbon of deoxyribose sugar, and a phosphate bound via ester linkage to the 5' carbon of the deoxyribose. DNA is composed of 2 antiparallel strands with opposite orientation of 5' terminal phosphate and 3' terminal hydroxyl. Each strand is a polymer of nucleotides in which nucleotides are covalently linked by a phosphodiester bond, in which the phosphate of one nucleotide is bound via an ester linkage to the hydroxyl attached to the 3' carbon of deoxyribose sugar of an adjacent nucleotide. The two complementary strands are held together by a hydrogen bonding between opposing, complimentary bases on opposing strands; adenine pairs by 2 hydrogen bonds with thymine; guanine pair by 3 hydrogen bonds with cytosine. Adjacent pairs in the "stack of base pairs are also held together weakly by van der Waals interactions. The two strands each coil to forma double helix. the angles of the glycosidic bonds between deoxyribose sugars and bases cause the sugar phosphate backbones to be brought closer together to each other than 180 degrees; as a result, bases are alternately more exposed in "major" grooves and less exposed and less exposed in "minor" grooves. DNA differs from RNA: DNA double-stranded vs. RNA single stranded DNA utilizes the nucleotide Thymidine; RNA utilizes uridine instead DNA nucleotides have the sugar deoxyribose; RNA nucleotides have the sugar ribose.
Address each of the following with respect to a chemical reaction: c. Describe how enzymes catalyze reactions with respect to activation energy and draw and explain a graph depicting this effect.
Enzymes catalyze reactions by lowering the activation energy required to reach the transition state, thus facilitating formation of the transition state and increasing the rate of the reaction. As shown in in graph, the effect of an enzyme is to lower the free energy of transition state, meaning that the input of energy to reach transition state, activation energy, is reduced by the enzyme.
3.2 Name the three amino acids that form the catalytic triad for chymotrypsin and describe the roles each of these amino acids pay in the catalysis of peptide bond hydrolysis.
Histidine 57: serves as a proton acceptor/donor when the polypeptide substrate enters the active site, histidine accepts a proton for the serine 195 in the enzyme active site, making histidine positively charged, making serine into a strong nucleophile. The accepted proton is later transferred from histidine to the free amino acid group at the cleaved end after peptide bond cleavage, allowing one of the peptides generated by cleavage to be released. Histidine then accepts a proton from water, allowing the remaining hydroxyl group to associate with the terminal carboxyl carbon of the other peptide fragments, allowing the fragment to be released from serine 195 this second accepted proton is then donated back from histidine to serine to re-establish serine's hydroxyl group of its side chain.
2.2. Also, describe how the relationship differences between DNA in a low ionic strength solution and DNA in a high ionic strength solution. Explain the reasons for this difference.
In a low ionic strength solution (low salt concentration) the negative charges of the phosphate groups on each of the two sugar-phosphate "backbones" of each strand causes the "backbones" to repel each other slightly, conferring a degree of instability on the double stranded DNA molecule, meaning less heat is required to separate the strands. When the DNA is in a higher ionic strength solution(high salt concentration) the positive ions in the solution will form electrostatic interactions with the negatively charged phosphate groups on the sugar-phosphate backbones of each DNA strand, masking of neutralizing these negative charges. This means the DNA molecule is more stable and requires more heat to dissociate the two strands.
Explain the purpose of Western blotting and describe the major steps involved in the technique.
Purpose: To determine whether a sample contains a specific protein of interest (determine where cells in a tissue sample express a specific protein)(OR to compare samples for their relative levels of a specific protein of interest) Steps: 1. Protein sample prepared in SDS to denature proteins and coat them in negative charge(lyses cells, if sample is cellular) 2. Run proteins inn sample is SDS/PAGE. Separates proteins in a gel based on differences in their molecular mass. 3.Transfer proteins from gel to a membrane/nitrocelluose 4.Block membrane(blocking buffer nonfat dry milk) to prevent non-specific binding of antibodies to membrane. 5. Probe blot with primary antibody that specifically binds protein of interest 6. Probe blot with secondary antibody that specifically binds primary antibody and carries a detectable label. 7. Run procedure for detecting label (colorimetric enzyme reaction, fluorescent molecule)
5.2 Describe the two techniques: co-immunoprecipitation and sandwich ELISA. Be sure to explain the main steps involved in each. Also, describe the purpose of each technique, i.e what type o information is provided by each technique.
Sandwich ELISA: Purpose: to detect presence of a specific protein in a sample and to quantitate relative levels of expressions to the protein in the sample (e.g as compared to another sample) Steps: 1. Wells in a plate (microtiter plate) are coated with an antibody specific for a protein of interest. 2.Sample(cell lysate) is pipetted into well allowing protein (if its in the sample) to bind to antibodies coating the well. 3. Non-bound portion of the sample is removed 4. A secondary antibody, also specific protein if interest( and that binds a different epitope) is added; this second antibody has a detectable label attached(e.g an enzyme or fluorophore with proper wavelength) 5. Intensity of signal is read to determine how much of the protein of interest was in the sample.
2.1.Draw a graph showing the relationship between melting temperature and Tm and G+C content of DNA and describe the underlying reasons for the relationship.
The melting temperature of DNA increases with increased G+C content since G-C base pairs have three hydrogen bonds and A-T base pairs only have two hydrogen bonds. As a result, the higher the G+C content of a DNA molecule, the more G-C pairs there are, and therefore more hydrogen bonds need to broken to "melt" the DNA i.e to dissociate the two strands. Breaking more hydrogen bonds require more heat and, therefore, a higher temperature.
Address each of the following with respect to a chemical reaction: a. Explain what "activation energy" is with respect to a chemical reaction.
a. Activation energy is the difference in free energy between reactant(s) (substrate(s)) and the transition state, the conformation between the reactant(s) and product(s). in other words, activation energy is the amount of energy that must be added to allow the reactant(s) to reach the transition state
5.1 Describe the two techniques: co-immunoprecipitation and sandwich ELISA. Be sure to explain the main steps involved in each. Also, describe the purpose of each technique, i.e what type o information is provided by each technique.
co-immunoprecipitation: Purpose: to determine which proteins normally bind to a protein of interest Steps: 1. Sample(e.g cell lysate) is mixed with beads to which antibodies for protein of interest are attached. 2.Sample is then centrifuged to cause beads, attached to the protein of interest, and any protein attached to protein of interest, to precipitate to the bottom of the tube forming a pellet of immunoprecipitated proteins attached to antibody on beads. 3.Supernatant(carrying the unbound proteins) is removed and discarded. 4.Pellet is washed to remove any unbound proteins. 5.Pellet is treated with denaturant to remove immunoprecipitated protein(s) from antibody. 6.Immunoprecipitated proteins can be analyzed by other techniques(e.g Western blotting, mass spectroscopy) to identify proteins that were immunoprecipitated because they were attached to protein of interest.
