The natural sulfur cycle involves many sulfur- reducing and sulfur-oxidizing bacteria. For example, S2- is oxidized to elemental sulfur and to [SO4]2-, and the reverse processes convert mobile [SO4]2- to immobilized S2-. (a). At the end of the sulfur-reduction sequence, HS- may be produced instead of S2-. What influences this outcome, and what other product is possible? (b). Sulfur-reducing bacteria can be applied to the removal of mobile heavy metals (e.g. Pb,Cd,Hg) from the environment. Explain why the metals are termed 'mobile', and describe the chemical processes and equilibria that lead to the immobilization of the metals.
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- The succinate dehydrogenase complex couples the oxidation of succinate to the reduction of ubiquinone (Q) according to the following equation. succinate + Q --> fumarate + QH2 Given that E'° for the fumarate/succinate redox pair is + 0.031V and E'° for the Q/QH2 redox pair is + 0.045V, calculate the standard free energy change (AG") for the oxidation of succinate by ubiquinone. 1º = - 1.35 kJ/mol Ag° = - 2.7 kJ/mol OAG° = - 280 kJ/mol OAG° = + 1.35 kJ/mol %3D OAG° = + 2.7 kJ/mol %3DIn the first step of the aldolase reaction, an active site Lys229 residue, with its side chain amino group in the deprotonated state, acts as a nucleophile and attacks the carbonyl C2 carbon of fructose 1,6-bisphosphate to form a Schiff base (boxed in the scheme). Since the pKa of the Lys side chain amino group in free solution is ~10.5, the pKa of Lys229 side chain must have been perturbed to a (higher lower) value for the enzyme to be active at neutral pH. the answer should include sufficient details, including the definition of pKa.Failure of anti-oxidant function results in the hydroxylation of an aromatic acid of Enzyme Z and its activation, so that it degrades protoporphyrin to porphyrin, an unstable product. When hit by light, this product further degrades to form a compound responsible for the lesions and excruciating pain the man suffers. The mutation also affected an amino acid at the N-terminal of Enzyme X. Sequencing of the first seven (7) amino acids at the N-terminal of the normal enzyme gave the following sequence: trp-arg-asp-leu-ser-gly-his When the CDNA was sequenced by the Sanger method utilizing ddCTP, the following products were obtained: Tetranucleotide, Hexanucleotide, Nonanucleotide, Decanucleotide, Dodenucleotide, Octadecanucleotide, Nonadecanucleotide, 21-nucleotide What is the sequence of the bases in the mRNA coding for the peptide above? The mutation involved the 19 bases of the template strand of the peptide. A comparison of the electrophoretic profile of the normal peptided (N) and and…
- Failure of anti-oxidant function results in the hydroxylation of an aromatic acid of Enzyme Z and its activation, so that it degrades protoporphyrin to porphyrin, an unstable product. When hit by light, this product further degrades to form a compound responsible for the lesions and excruciating pain the man suffers. The mutation also affected an amino acid at the N-terminal of Enzyme X. Sequencing of the first seven (7) amino acids at the N-terminal of the normal enzyme gave the following sequence: trp-arg-asp-leu-ser-gly-his When the cDNA was sequenced by the Sanger method utilizing ddCTP, the following products were obtained: Tetranucleotide Hexanucleotide Nonanucleotide Decanucleotide Dodenucleotide Octadecanucleotide Nonadecanucleotide 21-nucleotide The mutation involved the 19th bases of the template strand of the peptide. A comparison of the electrophoretic profile of the normal peptide (N) and mutant peptide (M) is shown below. The (+) electrode is situated at the bottom. pH…what cell types are Hexokinase I and Hexokinase IV found? Explain 2 additional differences (in addition to cell-type specificity) that exist between these two isoenzymes.From the complete oxidation of glucose (glucose → 6CO2), how many total nucleotide triphosphates are yielded (be sure to deduct payback) as part of substrate level phosphorylation?
- Failure of anti-oxidant function results in the hydroxylation of an aromatic acid of Enzyme Z and its activation, so that it degrades protoporphyrin to porphyrin, an unstable product. When hit by light, this product further degrades to form a compound responsible for the lesions and excruciating pain the man suffers. The mutation also affected an amino acid at the N-terminal of Enzyme X. Sequencing of the first seven (7) amino acids at the N-terminal of the normal enzyme gave the following sequence: trp-arg-asp-leu-ser-gly-his When the cDNA was sequenced by the Sanger method utilizing ddCTP, the following products were obtained: Tetranucleotide Hexanucleotide Nonanucleotide Decanucleotide Dodenucleotide Octadecanucleotide Nonadecanucleotide 21-nucleotide What is the sequence of the bases in the mRNA coding for the peptide above?Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form (dipole-dipole interactions, hydrodgen bonds, London Dispersion Interactions, or Ion-Dipole Interactions) with the hydroxyl groups of the glucose substrate and can potentially function as either a (hydrogen bond donor and/or acceptor, hydrogen bond donor, or hydrogen bond acceptor) . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the (increase or decrease)…In addition to the reactions mentioned in Section 23.5, PLP can catalyze b-substitution reactions. Propose a mechanism for the following PLP-catalyzedb-substitution reaction:
- Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form with the hydroxyl groups of the glucose substrate and can potentially function as either a . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the in glucose affinity as indicated by the higher KM for the mutant enzyme. The side chain of Asp could potentially serve as a , but…Some bacteria contain three different forms of aspartokinase, each with its own mode of regulation. Based on the roles of aspartoki- nase, as discussed in the text, propose a regulatory scheme applicable to each form of aspartokinase.A very new Hemoglobin variant, called Hb simplex, has just been identified in patients with hypoxia (i.e. patients for whom the oxygenation of organs is not properly accomplished). However, this variant is able to bind dioxygen with a Kd which is significantly lower to that of Hemoglobin from normal patients. This effect has been associated with the substitution of a Lys by a Met at two distinct sites of the b subunit of Hemoglobin : one site close to the N-terminus, and the other site close to the C-terminus of the polypeptide chain. Most importantly, these 2 sites are known to be involved in the maintain of the physical interactions between the subunits a and b in Hemoglobin. 1) The aforementioned Hb simplex mutations lead to a loss of physical interactions between the subunits a and b. What type(s) of chemical bonds between the 2 subunits (i.e. covalent, electrostatic, ion-dipole, dipole-dipole, van der waals) would you hypothesize to be affected by these mutations? explain. 2)…