We’re going to note which amino acid is most prevalent in the structure. We have the structure of the active site of a laccase above. Answer choice C is correct.ĥ) To answer this question, we can revisit the passage. That means our correct answer here is going to be 9 sp 2-hybridized carbon atoms present in AMC. There are 10 total carbon atoms in AMC, with 9 of these carbons having a single double bond. Sp 2-hybridized means a set of hybrid orbitals produced when one s orbital and two p orbitals are combined mathematically to form three new equivalent orbitals. From reading the passage we know we’re looking at AMC: (1 x 10 -4) / (2 x 10 -9) = 5 x 10 4 or answer choice D.Ĥ) First thing we’re going to do is identify the molecule in question. Dividing peptide concentration by proteasome concentration gives us Note in the experimental procedure we’re told the proteasome concentration of 2 nM (2 x 10 -9) while the peptide concentration is 100 μM (1 x 10 -4). We want to note the factor difference in concentration of the substrate and the concentration of the proteasome. In this case, the substrate is going to be the peptide succinyl–LLVY–AMC. We’re told a porphyrin compound can inhibit proteasome activity. The proteasome is the enzyme introduced by the author in the first part of the passage. We can pull up the relevant part of the passage to help us understand the question. ΔE = (6.62 × 10 –34) × (3.0 × 10 8) × which corresponds to answer choice C.ģ) This is a straightforward question in terms of how it’s worded, but we have to be careful to correctly identify the components mentioned in the question stem. We’ll plug the two wavelengths and our constants into our above equation to find the energy converted: We need to determine the energy converted based on the question stem, which will involve the two wavelengths I referenced from the passage. ![]() In this case, we have wavelengths so we can use the relationship between frequency and wavelength: f=c/λ. The energy of a photon is equal to the product of Planck’s constant and frequency of EM Wave: E = hf. We have fluorescence emission at 440 nm and excitation at 360 nm. First, what are the key points we need from the passage? Let’s revisit where the author talks about the excitation and emission events. ![]() Correct answer here is answer choice A.Ģ) This question is related to the passage, but ultimately, we’ll need to use external knowledge to deal with some math. In this specific question, we can find the amine in our structure: position I. ![]() Therefore, to break a peptide bond, it must be cleaved with the addition of a water, in a process called hydrolysis. During the formation of this peptide bond, the carboxyl (-COOH) group of one amino acid and the amino (-NH2) group of the other amino acid combine and release a molecule of water. Each amino acid is attached to its neighboring amino acid by a covalent bond, known as a peptide bond. What do we know about peptide bonds? We usually see them in amino acids. ![]() The author tells us AMC is liberated from a peptide as a result of peptide bond hydrolysis.ĪMC is attached to the peptide via a peptide bond. 1) To answer this question, we can go back to the passage and gather any necessary information.
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