MCAT Simulation Flashcards
Enzymes are heavily dependent on pH and temperature, and they function poorly outside their optimal conditions. High heat, for example, often denatures enzymes completely.
The “induced fit” model postulates that both enzyme and substrate change their conformations slightly to accommodate one another’s shape. This conformational change strengthens the binding between the two.
Allosteric inhibition involves the binding of a molecule at a site other than the active site. This binding causes a conformational change in the enzyme, rendering it unable to bind its original substrates. If the enzyme in the question stem is suddenly able to allosterically bind several new molecules, and those new substrates are present in the reaction environment along with the original substrate, then a significant proportion of the enzymes will be unable to carry out the original reaction.
Notice that proline’s side chain is actually bound to its amino group instead of projecting outward from the alpha carbon. Due to this rigid, cyclic side chain, proline forms specific angles when bound to other amino acids. For this reason, secondary structure (of which beta-pleated sheets are an example) would be affected.
Secondary structure includes the formation of alpha helices and beta-pleated sheets. Both of these structures are defined by patterns of hydrogen bonding.
All steroid hormones are derivatives of cholesterol. In general, steroid hormones can be distinguished from peptides by their suffixes, which tend to be either “-ol,” “-one,” or “-en.”
When naming a fatty acid, number the carbon atoms starting at the carbonyl. We can find the prefix from the total number of carbons in the chain (18 carbons = “octadec-”), while “-oic acid” is always used as the suffix. For unsaturated fatty acids, we must specify both the nature (cis vs. trans) and the location of the double bond(s); since all bonds span two carbons, remember to use the lower number. Finally, “-trien-” is inserted in between the prefix and suffix to designate the three alkenes shown
In eukaryotic cells in vivo, the role of cholesterol in the membrane is to provide fluidity within the otherwise rigid phospholipid structure.
In a saponification reaction, OH- acts as a nucleophile and attacks one of the carbonyl carbons to cleave the associated ester bond. Since one equivalent of NaOH is required to hydrolyze each ester, and since the question stem describes the freeing of two fatty acid chains, two equivalents are required.
Phospholipids are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. For this reason, a phospholipid will interact more favorably with water than will entirely hydrophobic molecules such as cholesterol or triacylglycerols. A positive (or less negative) change in entropy is a hallmark of a more favorable dissolution. In other words, since the solvation of a phospholipid is more favorable than the solvation of a cholesterol molecule, it will have a more positive entropy.
During the synthesis of malate, fumarate experiences the addition of water across its double bond. This results in the replacement of this bond with an –OH group on one carbon and a hydrogen atom on the other. Such a process represents a classic hydration reaction.
During the Krebs cycle, each unit of acetyl-CoA forms three molecules of NADH and one molecule of FADH2. In other words, the NADH-to-FADH2 ratio is 3:1. During a time period when nine molecules of NADH are produced, then, three units of FADH2 must also form.
These reactions actually occur during the course of the Krebs cycle. Specifically, they involve the enzyme isocitrate dehydrogenase. Isocitrate experiences the removal of protons to form a double bond, along with the loss of carbon dioxide; together, these processes form alpha-ketoglutarate.
Acetyl-CoA is fed into the Krebs cycle to produce three NADH, one FADH2, and one GTP molecule, the last of which is immediately converted to ATP. Each NADH is later used to form three ATP molecules, while the single FADH2 unit is utilized to synthesize two. In total, 12 molecules of ATP are created.
The TCA cycle is an alternate term for the Krebs cycle. Like most biological pathways, this process is inhibited by its own products in a classic form of negative feedback. Products of the Krebs cycle include ATP, citrate, and NADH.
At physiological pH, tryptophan (W) has a negatively-charged carboxyl group and a positively-charged amino group. In total, tryptophan thus has a net charge of 0 (I). Because W lacks an acidic or basic R group, it is a neutral amino acid (III). Specifically, the side chain of tryptophan is aromatic.
Ion-exchange chromatography separates compounds based on net charge. In this case, the stationary phase is negative, meaning that positive compounds will be retained in the column. In other words, these species will elute more slowly because they will interact more with the stationary phase. Therefore, we would expect the polypeptide that elutes first to contain fewer positively-charged residues than the one that elutes second. Arginine (R) is positively charged at physiological pH, so it is unlikely to be common in the polypeptide referenced.
Entropic penalties are incurred when amino acids are exposed to an environment in which they display poor solubility. The interior of a protein is generally held far from the aqueous environment outside it. In this interior, hydrophobic residues are favored, while hydrophilic amino acids incur a penalty. As glutamine, serine, and glutamic acid are all hydrophilic, choice C is the best answer.
Tertiary structure is driven by the tendency of hydrophobic residues to bury themselves inside the protein, away from aqueous environments, as well as side chain interactions such as hydrogen bonding and disulfide bridges. We can therefore expect that a substitution between amino acids that vary in polarity, charge, or sulfur content might change this level of structure. Of all the replacements listed, exchanging cysteine (a polar AA with the ability to participate in disulfide bridges) to glycine (a nonpolar, neutral residue that lacks sulfur) is most likely to cause a change in tertiary structure.
The table lists three pKa values. From outside knowledge, we should know that all amino acids possess at least one carboxyl and one amino group. Regardless of the identity of the R group, the pKas of these two termini are around 2 and 9-10.5, respectively. With this in mind, we can identify Group 1 as the carboxylic acid and Group 3 as the amine. This means that Group 2 must be the R group, or side chain.
