UWORLD MCAT Biochem

Donor and acceptor G,C,T,A

Adenine (A) and thymine (T), which each have 1 donor and 1 acceptor atom, form two hydrogen bonds between them (note that thymine has two potential acceptors in the form of carbonyls, but only one participates in Watson-Crick pairing). Guanine (G) and cytosine (C) pairs form three hydrogen bonds: Guanine contributes 1 acceptor and 2 donors whereas cytosine contributes 2 acceptors and 1 donor.

Prostaglandins

Educational objective: Prostaglandins are nonhydrolyzable, 20-carbon (eicosanoid) lipids involved in autocrine and paracrine signaling. They are derived from arachidonic acid and often help mediate localized inflammatory responses.

P186L mutation

Educational objective:RNA nucleotides are composed of a ribose sugar with a phosphate group bound to a nitrogenous base (eg, adenine, guanine, cytosine, and uracil). Adjacent sugars in the RNA backbone are bound to each other by phosphodiester bonds, and bases are linked to sugars by glycosidic bonds.

DNA denaturation and reannealing

The DNA double helix is arranged such that the sugar-phosphate backbone (linked by phosphodiester bonds) lies on the outside and the nitrogen-containing bases (paired by hydrogen bonds) lie on the inside. During double helix formation, complementary strands of DNA anneal through the creation of hydrogen bonds between the nitrogenous bases. Once the double helix is formed, energy is required to denature the duplex into two separate strands. Increasing the temperature during the melting step provides the heat energy necessary to break the hydrogen bonds between bases and separate the double helix. (Choice A) The base pairs in the double helix are "stacked" almost on top of each other; however, base stacking is interrupted, not enhanced, when the DNA helix denatures during melting. Stacking interactions, which are due to hydrophobic effects and nonpolar van der Waals forces between the bases, are primarily responsible for the stability of the double helix structure. (Choice B) Hydrophobic effects decrease during the melting step because hydrophobic bases are exposed to water. In aqueous solution, hydrophobic effects force hydrophobic bases toward the double helix interior and away from the polar solvent (water). In contrast, the hydrophilic, negatively charged phosphates in the backbone can hydrogen bond with water molecules. (Choice D) Covalent phosphodiester bonds of the sugar-phosphate backbone maintain the structure of single-stranded DNA. These linkages are not affected when the double helix is separated during denaturation. Educational objective:During DNA double helix formation, hydrogen bonds form between the bases and release energy. Heat energy provided in a DNA melting reaction is used to disrupt interactions that contribute to DNA double helix structure: hydrogen bonding, base stacking, and hydrophobic effects.

Cells communicate

Educational objective:Cells communicate through chemical signals called hormones. Hydrophilic hormones such as peptide hormones cannot cross the cell membrane and require second messengers. Hydrophobic hormones such as steroid hormones can cross the cell membrane and do not need second messengers.

Different amino acids within an enzyme affect the enzyme's activity to different extents

Educational objective:Different amino acids within an enzyme affect the enzyme's activity to different extents. Mutations to amino acids that are critical for proper folding or catalysis typically result in complete loss of activity whereas mutations in less critical amino acids result in small changes or no change in activity. The effects of a mutation must be determined experimentally.

Oncogenes

(Choice B) Inactivating mutations in tumor supressors can lead to cancer, but K-ras is a proto-oncogene. (Choices C and D) In cancer, the mutated K-ras protein has either normal function with increased expression or hyperactive (not hypoactive) function with normal expression. Educational objective:Oncogenes are mutated forms of genes involved in cell cycle progression (proto-oncogenes) that transform normal cells into cancer cells. Tumor suppressor genes regulate, pause, and inhibit cycle cell progression to ensure damaged DNA is repaired. In cancerous cells, mutations inactivate tumor suppressor genes and activate oncogenes.

Galactose

Monosaccharides are chains of three or more carbon atoms with the general formula CnH2nOn. Those with five or more carbons generally exist in equilibrium between cyclic and linear forms. In their linear forms, all monosaccharides contain one carbonyl group, and all other carbons in the chain are alcohols because they are bonded to OH groups. If the carbonyl is on one of the ends of the chain, it is an aldehyde. Monosaccharides with an aldehyde group are known as aldoses. Carbonyls within the carbon chain are ketones, and monosaccharides with a ketone functional group are known as ketoses. Galactose is an aldose, and as such it contains an aldehyde (Choice A) when released from GC and linearized. All other carbons in galactose are alcohols. The alcohols in the middle of the chain have one C-H bond, making them secondary alcohols (Choice D). The alcohol at the end of the chain has two C-H bonds, making it a primary alcohol (Choice B). Galactose is not a ketose, so it does not contain a ketone group. Educational objective:In their linear form, monosaccharides contain multiple alcohol groups and one carbonyl group. For aldoses, the carbonyl is an aldehyde, whereas for ketoses the carbonyl is a ketone. All monosaccharides contain at least one primary alcohol, and most contain at least one secondary alcohol.

A glycosidic bond is a bond between the anomeric carbon of a carbohydrate and any other biological molecule

A glycosidic bond is a bond between the anomeric carbon of a carbohydrate and any other biological molecule, including proteins, lipids, nucleotides, and other carbohydrates. A single carbohydrate can participate in multiple glycosidic bonds: First through its own anomeric carbon and then through bonds from its hydroxyl groups to the anomeric carbons of other sugars. The various ways in which carbohydrates can link to each other give rise to a high level of diversity among carbohydrate chains. Figure 1 shows that the D-galactose moiety of the A-antigen participates in three glycosidic bonds: One linking the anomeric carbon of D-galactose to the R-group, one linking the anomeric carbon of L-fucose to carbon 2 of D-galactose, and one between the anomeric carbon of GalNAc and carbon 3 of D-galactose. (Choice A) The terminal sugars GalNAc and L-fucose each participate in 1 glycosidic bond, but D-galactose participates in 3. (Choice B) The D-galactose moiety participates in 2 glycosidic bonds in the H-antigen, but not another glycosidic bond is added in the A-antigen. (Choice D) D-galactose would require another glycosidic bond at carbon 4 or carbon 6 to have a total of 4 glycosidic bonds. Educational objective:A glycosidic bond is a bond between the anomeric carbon of a carbohydrate and any other biomolecule. A single carbohydrate may participate in many glycosidic bonds by linking to the anomeric carbons of other carbohydrates, allowing a high level of diversity among carbohydrate chains.

Catecholamines

Catecholamines such as epinephrine and norepinephrine are hormones derived from the amino acid tyrosine. They can play a role in inflammatory responses but generally do so as endocrine signals (acting on cells that are distant from their origin).

Semiconservative DNA replication

DNA replication is a semiconservative process in which DNA polymerase synthesizes each daughter strand using one of the parental strands as a template. Each new double helix contains one parental strand and one daughter strand. In the experiment described, Escherichia coli bacteria initially have the heavy isotope 15N in their DNA. Once the bacteria are transferred to 14N media and allowed to replicate, newly synthesized strands will contain 14N. In Generation 1, all new strands will be paired with parental strands that contain 15N. Therefore, 100% of double helices produced in Generation 1 will have one 15N strand and one 14N strand (Choice D). In Generation 2, both the 15N and 14N strands from Generation 1 are used as templates for the synthesis of new 14N strands. In the DNA that used a 14N strand as a template, both parental and daughter strands will contain 14N nucleotides. In DNA that used a 15N strand as a template, the parental strand will contain 15N and the daughter strand will contain 14N. Because half of the parental strands are 15N and half are 14N, 50% of the Generation 2 double helices will contain one 15N strand and one 14N strand (Choice C). The other 50% will contain two 14N strands. In each new generation, the percentage of double helices containing one 15N strand and one 14N strand will be half the percentage of the previous generation. Therefore, in Generation 3, 25% of double helices will contain a 15N strand and a 14N strand. (Choice A) 12.5% of double helices will contain a 15N strand and a 14N strand in Generation 4, not Generation 3. Educational objective:DNA replication is a semiconservative process that results in each double helix containing one parental strand and one newly synthesized daughter strand. DNA polymerase synthesizes each new daughter strand by using a parental strand as a template.

histone acetylation

DNA winds tightly around proteins referred to as histones to form structural units known as nucleosomes. Gene expression depends partially on the association of histones with DNA. Genes that are actively transcribed are found on unwound stretches of chromatin called euchromatin, which transcription machinery can easily access. Inactive genes are usually in highly condensed DNA known as heterochromatin, which is much less accessible. Covalent post-translational modifications such as methylation, acetylation, and phosphorylation alter the association of histones with DNA. DNA is predominantly negatively charged due to the phosphate groups on the backbone. Histones associate with DNA by forming salt bridges between positively charged amino acid residues and negatively charged phosphate groups. These ionic interactions allow histones to bind tightly to DNA and prevent genes from being transcribed. Acetylation of histones involves the transfer of acetyl groups from acetyl coenzyme A to positively charged amino groups on lysine or arginine residues. This modification disrupts salt bridges by reducing the positive charge on histones, which allows DNA to unwind and become more accessible to transcription machinery. As a result, the acetylation of histones causes nucleosomes to relax and increases gene expression. (Choice A) Negatively charged amino acids (aspartate and glutamate) would be repelled by the negatively charged phosphate groups on DNA. (Choice B) The carboxyl oxygen atoms in the acetyl groups form hydrogen bonds with water rather than the nitrogenous bases on DNA. Acetylation neutralizes the positive charge on histones and decreases their interaction with DNA. (Choice D) Lysine residues are positively charged, and associate with negatively charged phosphate groups in DNA. If phosphate groups were positively charged, they would repel lysine. Educational objective:Histones associate with DNA by forming salt bridges between their positively charged residues and negatively charged phosphate groups on DNA. Acetylation of histones involves the transfer of an acetyl group to positively charged amino groups on lysine or arginine residues, increasing gene expression by disrupting salt bridges between histones and DNA.

Effect of increased enzyme concentration

Educational objective:An increase in the amount of enzyme available in a particular system will cause an increase in Vmax but no change in Km in Michaelis-Menten kinetics. Lineweaver-Burke graphs are linear plots that relate inverse reaction velocity (1/V) to inverse substrate concentration (1/[S]). An increase in enzyme concentration would cause a decrease in the y-intercept (1/Vmax) and no change in the x- intercept (−1/Km) in a Lineweaver-Burke plot.

Creating michaelis mentis plots

Educational objective:The Michaelis-Menten equation models the initial rates of a reaction at various substrate concentrations. Rates are measured as the slope of the initial, linear phase of the reaction before substrate is depleted and product accumulates. Enzyme concentration should be much lower than all substrate concentrations tested.

Cis or trans

Fatty acids are nonpolar molecules composed of straight hydrocarbon chains with carboxyl groups at one end. Humans synthesize fatty acids with an even number of carbon atoms, and chains usually range from 14 to 18 carbon atoms long. Fatty acids with no carbon-carbon double bonds are described as saturated because each carbon atom has the maximum number of hydrogen atoms possible. In contrast, an unsaturated fatty acidcontains one (monounsaturated) or multiple (polyunsaturated) double bonds that may be in either the cis (Z) or the trans (E) configuration. Unsaturation contributes significantly to membrane fluidity. The carbon-carbon double bonds decrease the melting temperature of fatty acid chains and increase the average space between lipids. As a result, lipids with unsaturated fatty acid chains remain liquid (fluid) at room temperature. The cis configuration is particularly important in cell membranes as it introduces a bend or "kink" in the fatty acid that prevents phospholipids from stacking together and solidifying. The passage introduces various molecules that have different effects on membrane fluidity. Of the choices listed, the cis bond of the polyunsaturated fatty acid EPA contributes the most to membrane fluidity. (Choice B) The trans bond in cer-EOS does not increase the membrane fluidity significantly as it does not disrupt the straight shape of the fatty acid chain. (Choices C and D) The carboxyl groups of free fatty acids such as DHA and the phosphate groups of phospholipids do not contribute to membrane fluidity. They are hydrophilic (polar) regions that associate with the aqueous sides of the cell membrane bilayer. Educational objective:Saturated fatty acids have no carbon-carbon double bonds and therefore stack closely in the cell membrane. Unsaturated fatty acids have one or more double bonds in either the cis or the trans configuration. In particular, the cis configuration enhances membrane fluidity by introducing a bend or "kink" in the fatty acid chain.

Ketone bodies

Ketone bodies are fuel sources that are produced under low glucose conditions. They are derived from acetyl-CoA and are not involved in inflammatory responses.

D or L

Molecules may have one or more stereocenters (an atom bonded to four unique substituents), each of which can adopt one of two stereochemical configurations with distinct properties. Every stereocenter within a molecule is given an R or S designation based on the relative positions of the substituents and their priority ranking. Priority is determined by the atomic mass of each atom to which the stereocenter is bonded, with higher mass given higher priority. If two or more of the bonds are to atoms of the same atomic mass, then the atoms to which they are bonded are assessed until a difference is found. With the exception of dihydroxyacetone, all carbohydrates have at least one stereocenter. Most monosaccharides (single sugar units) contain multiple stereocenters and cannot be designated as R or Sbecause R and S configurations refer only to individual stereocenters. Instead, each carbon is given a number starting at one end of the chain. The anomeric carbon (two bonds to oxygen) is given the lowest number possible, and the carbohydrate is then given an L or D designation based on the configuration of the stereocenter with the highest number. An R configuration at this position corresponds to a D-sugar, and an Sconfiguration corresponds to an L-sugar. Biological structures such as the myelin sheath contain L-amino acids and D-sugars almost exclusively. The galactose component of GC shown in Figure 1 has an R configuration at carbon 5, making it a D-sugar. (Choice A) L-Galactose is not used in biological structures. Figure 1 shows D-galactose. (Choices C and D) R and S designations are used to classify individual stereocenters within a molecule. D-Galactose has both R and S stereocenters, but the molecule as a whole cannot be classified as either R or S. Educational objective:Stereocenters are designated as having R or S configurations based on the arrangement of their substituents and the priority ranking of each. Carbohydrates with multiple stereocenters are given an L or D designation based on the configuration of the highest-numbered stereocenter. Almost all carbohydrates found in nature are in the D configuration.

DNA reannealing speeds

The four short tandem repeats sequences shown in Table 1 were denatured during melting curve analysis and cooled in environments with different pH and NaCl concentrations. Sequence 3 will have the fastest reannealing time as it is the shortest sequence and was allowed to cool in the solution closest to physiological pH with a high salt concentration (>0.15 M). (Choices A and B) Sequences 1 and 2 were studied in solutions with high pH and low salt concentration. Alkaline pH causes a loss of base pairing interactions, and low salt concentration would induce greater repulsion between negatively charged phosphate groups on DNA. In addition, sequence 2 is the longest and would take longer to reanneal. (Choice D) Sequence 4 DNA has a higher number of repeats than sequence 3 and is likely to be destabilized by the extremely acidic pH despite the high salt concentration. Educational objective:The following parameters influence the thermodynamic stability of the DNA duplex: DNA length: Longer DNA molecules take more time to both melt and reanneal. pH: Extreme changes in pH outside the physiological range lead to loss of hydrogen bonding and destabilize the DNA helix. Salt concentration (ionic strength): High salt concentration of the solution increases double helix stability, but decreased salt concentration decreases stability.

Phosphatidylserine

The plasma membrane is composed mainly of proteins, phospholipids, and cholesterol arranged in a bilayer with hydrophilic surfaces enclosing a hydrophobic interior. Phospholipids have a wide range of structures but generally consist of one or more fatty acid tails linked to a polar head group that contains a phosphate. The charges of phospholipids depend on the head groups attached to the phosphate. Head groups can be either neutral or negatively charged. Influenza A obtains its viral envelope from the plasma membrane of its host cell. However, the passage states that the composition of the envelope differs significantly from the plasma membrane. Unlike human membranes, the viral envelope contains mostly phosphatidylethanolamine and negatively charged phospholipids. Therefore, phospholipids with ethanolamineand negatively charged head groups such as serine, inositol, and diphosphatidylglycerol are needed for envelope formation. Of the choices presented, phosphatidylserine is the only negatively charged phospholipid. Therefore, its absence is most likely to affect the assembly of the viral envelope. (Choice A) Palmitic acid is the only free fatty acid synthesized by cells. Although it can be incorporated into phospholipids, the passage does not indicate that it is required. (Choice B) Triglycerides are storage lipids found in adipocytes (fat cells) and consist of three fatty acid tails bound to a glycerol backbone by ester linkages. They are not found in cell membranes or viral envelopes. (Choice C) Sphingomyelin is a phospholipid with a sphingosine backbone. The phospholipids required for the viral envelope (phosphatidylethanolamine and negatively charged glycerophospholipids) do not contain sphingosine, but instead contain a glycerol backbone. Sphingomyelins are not part of the viral envelope. Educational objective:Phospholipids are composed of a hydrophilic polar head that contains a phosphate group and a hydrophobic tail with one or more fatty acids. They may contain a glycerol or a sphingosine backbone, and their properties depend on the head groups attached to the phosphate.

Sigmoidal curves indicate enzymes with multiple, cooperative active sites

The rate of an enzyme-catalyzed reaction depends on the concentration of substrate. For enzymes with a single active site, the reaction rate increases steadily with substrate concentration until the active sites become saturated. Once saturation is achieved, adding more substrate does not increase the reaction rate. This takes the form of a hyperbolic curve when reaction rate is plotted as a function of substrate concentration. Some enzymes have multiple active sites and can bind more than one substrate at a time. Many of these enzymes behave cooperatively such that substrate binding at one active site alters the affinity of other active sites for the substrate. In positive cooperativity, binding at one active site increases the affinity, and therefore the reaction rate, of other active sites. This results in an "S"-shaped curve, called a sigmoidal curve, in which the rate increases slowly at first, increases more quickly as the affinity for substrate increases, and finally levels off as all active sites become saturated. Only enzymes with multiple active sites can exhibit cooperativity. Because it shows a sigmoidal curve, enzyme 1 has positive cooperativity and therefore must have multiple active sites. (Choices B and C) Enzyme 2 has a hyperbolic curve, and therefore does not exhibit cooperativity. Although it is possible for a noncooperative enzyme to have multiple active sites, it is not required. (Choice D) Enzyme 1 exhibits positive cooperativity, and therefore must have more than one active site. Educational objective:Enzymes that display positive cooperativity yield sigmoidal curves instead of hyperbolic curves when reaction rate is plotted against substrate concentration. Because cooperativity is a change in the affinity of one active site for its substrate when another active site is bound, any enzyme that exhibits cooperativity must have more than one active site.

Mannosidase

According to the passage, D-mannose is the first sugar in the glycan chain and links to Dg via a glycosidic bond through the anomeric carbon of D-mannose. Mannosidase is in a class of enzymes called glycosidases that break glycosidic bonds. In this case, the broken bond is the bond that links the glycan chain to Dg through D-mannose, so the entire glycan chain will be removed from Dg as a single polysaccharide. The 32P is attached to carbon 6 of mannose, and the 3H ribitol 5-phosphate is within the cleaved polysaccharide. Therefore, both radiolabels will be removed from the Dg protein and remain on the cleaved polysaccharide. (Choice A) The polysaccharide would have to be cleaved between the ribitol 5-phosphate added by fukutin and the end of the glycan chain for both radiolabels to remain associated with Dg. Instead, mannosidase cleaves the bond that links D-mannose to Dg and removes all radiolables from Dg. (Choice B) If the polysaccharide were cleaved between mannose and ribitol 5-phosphate, this would be the correct answer. Because the bond linking the polysaccharide to Dg is cleaved, both radiolabels are removed from Dg. (Choice C) For this option to be correct, the polysaccharide would have to be cleaved in such a way as to remove mannose from Dg without removing ribitol 5-phosphate. Because mannose is the only sugar directly linked to Dg, this is not possible. Educational objective:Glycosidases break glycosidic bonds to produce monosaccharides, disaccharides, and polysaccharides. The length and composition of the newly formed saccharides depends on which glycosidic bond the glycosidase breaks.

Expresses enzyme

According to the passage, GalNAcyl transferase modifies the H-antigen by adding GalNAc in an α-1,3 glycosidic linkage to D-galactose, producing the A-antigen. Therefore, individuals who express the A-antigen on their red blood cells must express at least one copy of GalNAcyl transferase. This includes participants in the study with type A blood and those with type AB blood. Table 2 shows that 2746 participants had type A blood and 174 had type AB blood, for a total of 2920 participants who expressed the A-antigen. Therefore, 2920 participants expressed GalNAcyl transferase. The total number of participants was 10,000, so approximately 29% of the participants in the study expressed GalNAcyl transferase. (Choice B) Approximately 27% of participants had type A blood (2746/10,000), but individuals with type AB blood also express GalNAcyl transferase and must be counted in the calculation. (Choice C) Those with type B blood do not express GalNAcyl transferase. Approximately 11% of the participants had either type B blood or type AB blood (887+174/10,000). This represents the percentage of individuals who express galactosyl transferase. (Choice D) Approximately 9% of participants had type B blood (887/10,000). These individuals express galactosyl transferase, not GalNAcyl transferase. Educational objective:To determine the percentage of a population that meet specific criteria, all individuals with the relevant trait must be considered out of the total population evaluated.

glycosidic bond

All monosaccharides have one carbon atom that has two bonds to oxygen, known as the anomeric carbon. The chain-terminating carbon that is closest to the anomeric carbon is designated as position 1, and all other carbons are numbered sequentially. In their linear form, aldoses such as galactose contain an aldehyde group. Because aldehydes are always at the end of a carbon chain (chain-terminating groups), the anomeric carbon of an aldose is always carbon 1. Ketoses contain ketone groups, which are not chain-terminating. Although a ketose could theoretically have a carbonyl at any non-chain-terminating carbon, biologically important ketoses(eg, fructose) have their anomeric carbon at carbon 2. A glycosidic bond links the anomeric carbon of a carbohydrate to another molecule such as a lipid, a protein, a nitrogen base, or another carbohydrate. According to the passage, GalC hydrolyzes a glycosidic bond in GC to release the carbohydrate galactose and the lipid ceramide. Therefore, GalC cleaves the bond between carbon 1 of galactose and an oxygen atom of ceramide. (Choice B) The bond between carbon 1 of galactose and the oxygen atom in the galactose ring links galactose to itself, not to another molecule. It is not the glycosidic bond. (Choice C) The oxygen atom in the ring of galactose does not link galactose to another molecule. In addition, carbon 4 is not bonded to the ring oxygen. (Choice D) Carbon 4 is not the anomeric carbon and does not participate in the glycosidic bond. Educational objective:Glycosidic bonds link the anomeric carbon of a carbohydrate to another molecule, such as a lipid, a protein, a nitrogen base, or another carbohydrate. The anomeric carbon has two bonds to oxygen, and for aldoses it is carbon 1. For biologically important ketoses, the anomeric carbon is carbon 2.

Western blot

Autoradiography only detects molecules that contain radioactive atoms; lanes containing molecules with no radiolabel will appear empty. However, it is also possible that those lanes appear empty because they contain no molecule at all, with or without radiolabels. The western blot was performed with an anti-Dg antibody that detects all Dg molecules and confirms that they are present, even if they are not radiolabeled. The researchers can then be sure that the absence of bands in the autoradiogram is due to failure of the relevant enzyme to modify Dg and not due to a problem with isolating Dg itself. Note that unmodified Dg runs further on the gel because the absence of carbohydrates yields a smaller molecule. Tests of this nature are called positive controls, and they help ensure that each step of the experimental procedure performs as expected. The first lane in each autoradiogram is also a positive control, showing that radiolabels are added when all enzymes are present. (Choice B) The dependent variable in an experiment is the parameter that may change in response to changes in experimental conditions. In this case, whether or not Dg gains a radiolabel (either 32P or 3H) is the dependent variable. The molecular weight of Dg decreases when enzymes are absent. (Choice C) The independent variable is the experimental condition that is changed by the researchers for each part of the experiment. In this case, the presence or absence of POMK and fukutin is the independent variable. (Choice D) A negative control shows what the outcome would look like if the results were negative and helps confirm that positive results come from the expected source. In this case, the lane with no POMK in the 32P experiment and the lane with no fukutin in the 3H experiment are negative controls because it is expected that the radiolabels will not be added when the relevant enzymes are absent. Educational objective:A proper experimental approach requires controls to confirm that the results are related to the conditions being tested rather than experimental error or confounding factors. Controls can be positive or negative, showing what a positive or a negative result should look like, respectively.

Enzyme substrate binding

Educational objective:Some enzymes can bind different substrates that have the same chemical bonds or linkages. The lock and key theory suggests that substrates bind readily to enzymes because their active sites are already in the correct structural conformation. The induced fit theory states that substrate binding induces a change in the shape of the active site to accommodate the substrate.

Catalytic turnover and catalytic efficiency

Catalytic turnover refers to the rate at which an enzyme-bound substrate is converted to product. It is given by the rate constant kcat and is proportional to the maximum reaction rate Vmax. On a Michaelis-Menten plot, catalytic turnover is reflected in the height of the plateau that occurs at high substrate concentrations. A high kcat(or a high plateau) corresponds to high catalytic turnover. Catalytic efficiency is defined by the ratio kcatKmkcatKm and is a measure of enzyme specificity. It indicates the rate of catalysis at low substrate concentrations and is proportional to the initial slope of the Michaelis-Menten plot. An enzyme that quickly creates product at low substrate concentrations (high initial slope) is said to be catalytically efficient because only a small amount of substrate is needed to generate substantial activity. A high kcat, a low Km, or both may contribute to catalytic efficiency. An enzyme-substrate system may have any combination of high and low efficiency and turnover. In this case, CSL-173 has the highest kcat (turnover) and CSL-176 has the highest kcatKmkcatKm ratio (efficiency). (Choice A) Although CSL-173 has the highest catalytic turnover (largest kcat), WKLL-ACC has the lowest catalytic efficiency ( kcatKmkcatKm ratio). (Choice B) CSL-176 is the most efficient substrate, but WKLL-ACC has the lowest turnover. (Choice C) This option has the optimal catalytic turnover and catalytic efficiency reversed. Educational objective:Catalytic turnover is represented by kcat and is proportional to the maximum reaction velocity, which is obtained at high substrate concentrations. Catalytic efficiency is represented by kcatKmkcatKm and reflects the ability of the enzyme to work at low substrate concentrations.

Nucleotide structure

Deoxyribonucleotides are the monomer units that make up the structure of DNA. A nucleotide is composed of a nucleoside molecule attached by a phosphoester bond to one or more phosphate groups. Nucleosides consist of a pentose (five-carbon) sugar linked to a nitrogenous base on the 1′ carbon by a covalent glycosidic bond. Ribose and deoxyribose are the two types of pentose sugars used in nucleotide production, and only differ in structure at the 2′ carbon. Ribose sugars have a hydroxyl group at the 2′ carbon where deoxyribose sugars have a hydrogen atom. DNA replication only utilizes nucleotides that contain deoxyribose sugars. To create dNTP derivatives of dFTP, researchers substituted thymine bases with difluorotoluene. Because the remaining nucleotide structure is unaltered, dFTP nucleotides would not contain hydroxyl groups on the 2′ carbon of the pentose sugar. (Choice A) Deoxyribonucleotides, including those made with dFTP, all contain phosphate groups on the 5′ carbon of the pentose sugar. (Choice B) All nucleotides and nucleosides are bonded to nitrogenous bases at the 1′ carbon of the pentose sugar through a glycosidic bond, which is a bond between a carbohydrate and any other group. (Choice D) The H group on the 5′ carbon of the pentose sugar is present in nucleosides and nucleotides. Educational objective:A nucleoside is a pentose (five-carbon) sugar linked to a nitrogenous base on the 1′ carbon by a covalent glycosidic bond. Nucleotides consist of a nucleoside attached to one or more phosphate groups by a phosphoester bond. If the pentose has a hydroxyl group at the 2′ carbon, it is ribose; if a hydrogen is present at the same position, it is deoxyribose.

Effect of Kcat and Km on catalytic effieciency

Educational objective:Catalytic efficiency is a measure of an enzyme's ability to catalyze reactions at low substrate concentrations. Efficiency is calculated as kcat/Km, and its value can be increased by increasing kcat or by decreasing Km.

Fischer projections are two-dimensional representations of molecules that are often used to distinguish among carbohydrate anomers, epimers, and enantiomers.

Educational objective:Fischer projections help distinguish anomers, epimers, and enantiomers. Enantiomers are compounds with the same chemical formula that differ at every stereocenter, whereas epimers and anomers differ at only one stereocenter. Anomers can exist only in cyclized sugars and differ at the anomeric carbon.

Melting points

Educational objective:Increasing temperatures can cause proteins to unfold (denature) by disrupting the intermolecular forces that maintain secondary, tertiary, and quaternary structures. When the melting temperature of a protein is achieved, there are equal amounts of folded and unfolded proteins. A lower melting temperature indicates lower stability and a higher denaturation rate.

Melting temperature Tm measures protein stability

Educational objective:Protein stability is determined by the strength of the intramolecular forces that maintain a protein in its folded form. The temperature at which 50% of proteins in solution become denatured (melting temperature Tm) is a measure of protein stability. A higher Tm corresponds to a more stable protein.

Glycoprotein

Educational objective:Reducing sugars contain free anomeric carbons that provide reducing power when they are oxidized. In linear form the anomeric carbon is an aldehyde or a ketone, and in cyclic form reducing sugars have hemiacetal or hemiketal configurations. Nonreducing sugars contain acetal or ketal structures in their cyclic forms.

Enatiomers

Enantiomers are molecules with the same molecular formula that are mirror images because they differ in the orientation of every stereocenter. In planar diagrams of sugars, the orientation of stereocenters is generally indicated by bonds that extend above (wedge bonds) or below (dash bonds) the plane of the rings. The anomeric carbon may adopt either configuration through mutarotation. The enantiomer of a D-sugar is the L-form of the same sugar. For example, the enantiomer of D-mannose is L-mannose. Most biologically relevant sugars adopt the D-configuration instead of the L-configuration, with fucose being a rare exception. Figure 1 shows the structure of L-fucose as part of the H-antigen. L-fucose is dehydroxylated at carbon 6 and is connected to substituents by dash bonds at carbons 3, 4, and 5 and wedge bonds at carbons 1 and 2. The anomeric carbon (carbon 1) is ignored, so the correct sugar should have the opposite stereochemistry at carbons 2, 3, 4, and 5. D-galactose contains a hydroxyl group at carbon 6, wedge bonds at carbons 3, 4, and 5, and a dash bond at carbon 2. Therefore, D-galactose has the opposite stereochemistry of L-fucose at every stereocenter excluding carbon 1. For this reason, L-fucose is also known as 6-deoxy-L-galactose and is the dehydroxylated enantiomer of D-galactose. (Choices A and B) Xylulose and ribulose are pentoses (five-carbon sugars) formed during the pentose phosphate pathway. L-Fucose is a dehydroxylated hexose (six carbons), so it cannot be the enantiomer of a pentose. In addition, L-xylulose can only be the enantiomer of a D-sugar. (Choice C) The fucose shown in Figure 1 is in the L-configuration, so an L-sugar such as L-glucose cannot be the enantiomer. Educational objective:Enantiomers are molecules with identical molecular formulae that differ in the configuration of every stereocenter. They are mirror images of each other. The enantiomer of a D-sugar is the L-form of the same sugar.

Catalysts

Enzymes are biological catalysts that increase the rate of metabolic reactions. Phosphoglycerate mutase is an enzyme that catalyzes the reversible conversion of 3-phosphoglycerate (3-PG) to 2-phosphoglycerate (2-PG). The forward reaction (3-PG to 2-PG) occurs in glycolysis, and the reverse reaction (2-PG to 3-PG) occurs in gluconeogenesis. Catalysts increase reaction rates by facilitating formation of a more stable transition state between reactants and products. This enhanced stability decreases the activation energy Ea necessary for the molecules to react, allowing them to reach the energy threshold more often at any given temperature. As a result, the reaction rate increases in both the forward and reverse directions, resulting in an increased rate constant as calculated by the Arrhenius equation (Choices A, C, and D). Catalysts only stabilize the transition state of a chemical reaction. They do not stabilize the reactants or products themselves, and therefore the difference in energy between them (ΔG°) is not affected. The equilibrium constant of a reaction Keq is the ratio of products to reactants at equilibrium, and is related to the standard free energy change by the equation ΔG° = −RT ln(Keq). Because ΔG° does not change in the presence of a catalyst, the equilibrium concentrations of products and reactants also do not change. Therefore, the addition of phosphoglycerate mutase does not change the equilibrium concentrations of 2-PG or 3-PG. Educational objective:Catalysts, including enzymes, increase reaction rate constants by stabilizing transition states and decreasing activation energy. They do not affect the energy states of products or reactants, and therefore they do not affect the equilibrium positions of the reactions they catalyze.

Kinase and phosphotase

Enzymes are divided into classes and subclasses based on the kinds of reactions they catalyze. Biological processes are frequently regulated by opposing reactions such as phosphorylation and dephosphorylation, with each reaction counteracting the other. Phosphorylation reactions are most commonly carried out by kinases, a type of transferase that moves a phosphate group (HPO42−) from a donor molecule (usually ATP) to an acceptor molecule. Dephosphorylation reactions are catalyzed by phosphatases that remove an inorganic phosphate group by adding water across a bond (hydrolysis). Reaction 1 shows that LCA enzymes catalyze the hydrolysis of the phosphoester bond in phosphotyrosine residues. Enzymes that hydrolyze bonds are known as hydrolases, and the subset of hydrolases that remove phosphate groups, including LCA enzymes, are classified as phosphatases. The question asks which type of enzymatic activity would most likely counteract the effects of LCA enzymes. Because LCA enzymes are phosphatases that facilitate the removal of phosphate groups, their activity is most likely counteracted by kinase activity, which adds phosphate. (Choice A) Dehydrogenases are a type of oxidoreductase that transfer electrons through hydrogen atoms or hydride ions. They do not counteract phosphatase activity. (Choice C) Synthetases are a type of ligase that link two molecules together and typically use ATP hydrolysis as a source of energy. They should not be confused with synthases, which are lyases and do not require energy input. Neither synthetases nor synthases phosphorylate acceptor molecules. (Choice D) Mutases are a type of isomerase that rearrange functional groups within a molecule. They do not counteract phosphatase activity. Educational objective:Phosphatases catalyze the dephosphorylation of their substrates by hydrolysis, producing inorganic phosphate. Kinases catalyze the phosphorylation of their substrates by transferring organic phosphate from a donor molecule, usually ATP. The two enzyme types often regulate biological processes by counteracting each others' effects.

melting curve

Figure 2 shows the DNA melting curve for wild-type (WT) K-ras DNA and mutant K-ras DNA produced by SW-480 cells. To generate a melting curve, identical copies of each type of double-stranded DNA are gradually exposed to increasing temperatures, which causes the separation (denaturation) of the double helix into two strands of DNA. The melting temperature (Tm), depicted as the midpoint of the melting curve, is achieved when 50% of double-stranded DNA has become single-stranded (ie, when half of the helical structure is lost). LC Green Plus is a compound that fluoresces strongly when bound to double-stranded DNA but has minimal fluorescence with single-stranded DNA. The amount of double-stranded DNA in a sample can be detected by measuring the fluorescence of LC Green Plus. Consequently, as temperature increases in DNA melting curves, the fluorescence decreases as double-stranded DNA becomes single-stranded.In Figure 2, both WT K-ras and SW-480 K-ras DNA have a maximal fluorescence of 0.04 at 20°C, which signals that all DNA molecules are double-stranded and bound by LC Green Plus. The fluorescence of WT K-ras DNA decreases to 0.02 at 55°C, which is the Tm of WT K-ras DNA and signifies that 50% of the WT DNA has melted and become single-stranded while the other 50% is double-stranded. However, mutant K-ras DNA from SW-480 cells reacts differently at 55°C; the fluorescence decreases to a value near zero, indicating that 100% of mutant K-ras DNA is single-stranded. (Choices A, C, and D) At at 55°C, Figure 2 shows that 100% of SW-480 K-ras DNA is single-stranded (not double-stranded) but that 50% (not 100%) of WT DNA is single-stranded.Educational objective:To produce a melting curve, double-stranded DNA is gradually exposed to increasing temperatures, which cause the separation of the double helix into two strands of DNA. The melting temperature (Tm) is achieved when 50% of double-stranded DNA has become single-stranded; Tm is usually depicted as the midpoint of the melting curve.

dNMP molecular weight

Gel electrophoresis separates molecules by molecular weight, with larger molecules migrating more slowly than smaller molecules. DNA is composed of the four deoxyribonucleotides (dNMPs), each of which has a different molecular weight. The purine deoxyguanosine monophosphate (dGMP) is the largest dNMP, followed by deoxyadenosine monophosphate (dAMP), deoxythymidine monophosphate (dTMP), and deoxycytidine monophosphate (dCMP). An oligonucleotide is a short strand of DNA, and its molecular weight is determined by its composition of dNMPs. An oligonucleotide composed entirely of dGMP will be larger than any other oligonucleotide of the same length and therefore would move the most slowly through an agarose gel. (Choices A, B, and C) Deoxyguanosine has the largest molecular weight of any nucleotide, so all other nucleotides will move faster through an agarose gel. Educational objective:DNA is a polymer composed entirely of deoxyribonucleoside monophosphate (dNMP) monomers. Ribonucleotides are not incorporated into replicated DNA. The molecular weight of dNMPs in decreasing order is deoxyguanosine (dGMP), deoxyadenosine (dAMP), deoxythymidine (dTMP), and deoxycytidine (dCMP).

Analog bases

In DNA replication, dNTPs containing thymine (dTTP) are selectively inserted opposite adenine nucleotides rather than guanine, cytosine, or other thymine nucleotides in the template strand. Scientists generally attribute this selectivity to the hydrogen bonding between thymine and adenine in the Watson-Crick model of DNA. The study described in the passage tested this hypothesis by synthesizing dNTP derivatives of dFTP, an analog that lacks Watson-Crick hydrogen bonding capacity but is similar in size and shape to thymine. The effect of hydrogen bonding on base selectivity was measured as the efficiency with which dTTP and dFTP were inserted into a growing primer strand by KF. Scientists predicted that only dTTP nucleotides would be selectively paired with adenine and dFTP would not be added to the growing strand. However, the results depicted in Figure 1 show no significant difference in the base selectivity between dTTP and its structural analog dFTP. When KF detected adenine in the template strand, dFTP was inserted with the same efficiency as dTTP. Similarly, when the template base was guanine, cytosine, or thymine, the efficiency of dFTP and dTTP insertion was greatly reduced. Therefore, dFTP exhibited the same base pair selectivity as dTTP despite its inability to form Watson-Crick hydrogen bonds. (Choice A) dFTP was efficiently inserted into DNA strands opposite adenine; therefore, it serves as a good substrate for DNA polymerase. (Choice C) The template strand in the experiment did not contain dFTP nucleotides, so the results are neither consistent nor inconsistent with the hypothesis that dFTP is an effective template nucleotide. (Choice D) dFTP is similar in shape to dTTP. Its selective incorporation into DNA suggests that shape is involved in selectivity. Educational objective:DNA polymerase selectively inserts complementary nucleotides that form hydrogen bonds with bases on the template strand. The Watson-Crick model shows that adenine is paired with thymine whereas guanine is paired with cytosine.

Cholesterol

Membrane fluidity is determined partially by the concentration of cholesterol and the tail length of fatty acids. Cholesterol is a steroid alcohol that maintains membrane fluidity at an optimal level. At high temperatures, it provides rigidity and stabilizes the membrane; at low temperatures, it increases fluidity and prevents the membrane from solidifying. Short unsaturated fatty acid tails increase membrane fluidity by preventing phospholipids from clustering together, but longer saturated tails such as those typically found in sphingolipids induce lipid clustering and decrease fluidity. As described in the passage, changes in cholesterol levels and sphingolipid concentration can lead to diseases such as atherosclerosis and psoriasis, respectively. Atherosclerosis results from the accumulation of cholesterol into CCDs in the membranes of endothelial cells. Free fatty acids such as O3FA can inhibit CCD formation by separating cholesterol molecules. Therefore, an increase in free fatty acids would help treat atherosclerosis. Psoriasis is marked by an increase in membrane permeability due to a lack of ceramides in the cell membranes of the SC. The long fatty acid chains in sphingolipids such as cer-EOS cause lipids to cluster and decrease permeability. Therefore, increased sphingolipid levels can help treat psoriasis. (Choice A) Decreasing LDL would improve atherosclerosis, but ceramides should increase, not decrease, in psoriasis therapies. (Choice B) Glycolipids are not involved in atherosclerosis, so decreasing their levels would not affect atherosclerosis. According to the passage, ceramides (a type of lipid derived from sphingosine), not cholesterol, are needed to prevent psoriasis. (Choice D) Terpenes are precursors in cholesterol synthesis, so increased terpene concentration could worsen atherosclerosis by increasing cholesterol production. Saturated lipids such as cer-EOS are necessary for psoriasis treatments. Educational objective:Cholesterol levels and fatty acid tail length help control membrane fluidity by regulating rigidity and lipid clustering. Changes in levels of either can result in disease.

autoradiography

In autoradiography, potentially radiolabeled biomolecules, including proteins, nucleic acids, and carbohydrates, are separated in gels. Radioactivity is then detected by exposure to radiation-sensitive film, and molecules that contain radioactive atoms such as 32P or 3H show up as bands. Molecules without radioactive atoms do not produce visible bands. Figure 2 shows that 32P from labeled ATP is incorporated into Dg in all cases tested except when POMK is absent. POMK is a kinase, meaning that it transfers a phosphate from ATP to a target molecule, so the lack of 32P-incorporation in the absence of POMK is expected. Similarly, 3H from ribitol 5-phosphate is not incorporated when fukutin is absent, which is expected because fukutin attaches ribitol 5-phosphate to the growing glycan chain. However, the figure also shows that in the presence of fukutin, ribitol 5-phosphate is incorporated into Dg when POMK is present but not when POMK is absent. Therefore, POMK activity must occur before fukutin can catalyze the addition of ribitol 5-phosphate to the glycan, and must occur first. (Choice A) If fukutin activity needed to occur before POMK activity could, no 32P-labeled band would appear in the lane where fukutin is absent. (Choice C) If fukutin and POMK activities could occur in any order, fukutin would be active in the second lane of the 3H autoradiogram, and a band would be visible. (Choice D) If POMK and fukutin activities needed to occur simultaneously, neither would show activity in the absence of the other and only the lanes with both enzymes would have visible bands in the autoradiogram. Educational objective:Autoradiography can detect the incorporation of radioactive atoms into biomolecules. It can help determine the conditions under which an enzyme can act on its substrate.

Lipid processing and classification

Lipid processing begins in the small intestine (duodenum), where bile salts break down lipid globules into smaller droplets in a process called emulsification. This process results in the formation of spherical structures, known as micelles, composed of a hydrophobic core containing the nonpolar hydrocarbon tails of lipids and an outer shell of polar head groups that make contact with water. The formation of micelles increases the surface area of lipid available for hydrolysis by lipases. Lipases are enzymes that digest certain emulsified lipids to facilitate their absorption, although some lipids are nonhydrolyzable. Hydrolyzable lipids contain ester bonds that can be cleaved by lipases through the addition of a water molecule (hydrolysis). These lipids include triacylglycerols, phospholipids, sphingolipids, and waxes. Nonhydrolyzable lipids do not contain the ester linkages necessary for lipase digestion. The most common nonhydrolyzable dietary lipids are cholesterol (steroids) and fat-soluble vitamins (A, D, E, and K). The question states that some dietary lipids are catabolized (broken down) by lipases, whereas others can be absorbed directly. The most accurate description of lipid processing acknowledges that all lipids are emulsified during digestion but only some lipids, including triglycerides and phospholipids, contain hydrolyzable ester bonds. (Choice A) Waxes are hydrolyzable lipids that consist of fatty acid chains bound to long-chain alcohols by an ester bond. (Choices B and C) All lipids are emulsified in the aqueous environment of the small intestine. Prostaglandinsare nonhydrolyzable lipids that function as signaling molecules in the human body. Educational objective:Emulsification increases the surface area of lipids by breaking down large globules into spherical structures called micelles. Micelles have a hydrophobic core, which contains the nonpolar hydrocarbon tails of lipids, and an outer shell of polar head groups, which make contact with water. Lipases can cleave the ester bonds in hydrolyzable lipids such as triglycerides, phospholipids, and waxes by adding a water molecule (hydrolysis).

Hydrolyzable lipids

Lipids are a broad class of hydrophobic biomolecules that are categorized at multiple levels. At a general level, they are classified as hydrolyzable or nonhydrolyzable based on whether they contain hydrolyzable linkages such as esters or amides. Hydrolyzable lipids are further classified based on their backbone structure and modifications. For example, lipids built on a glycerol backbone are glycerolipids, whereas those built on sphingosine backbones are sphingolipids. A lipid with a phosphate group modification is called a phospholipid, and a lipid with a carbohydrate attached is called a glycolipid. Multiple characteristics may be present at once, yielding more complex names such as glycerophospholipids, phosophosphingolipids, glycosphingolipids, or phosphoglycolipids. The passage states that GC is a hydrolyzable lipid. The structure in Figure 1 shows that it is built on a sphingosine backbone, making it a sphingolipid. Figure 1 and the passage also show that GC includes the carbohydrate galactose, attached to the lipid by a glycosidic bond (glycosylation). Because GC is a sphingolipid that has been glycosylated, it can be classified as a glycosphingolipid. (Choice A) Glycerophospholipids contain a phosphate group and are built on a glycerol backbone. Neither is true of galactocerebroside, so it cannot be classified as a glycerophospholipid. (Choices B and D) Galactocerebroside does not contain a phosphate group, so it cannot be classified as a phosphosphingolipid or a phosphoglycolipid. Educational objective:Hydrolyzable lipids are classified according to their modifications. Those with sphingosine backbones are sphingolipids; those with glycerol backbones are glycerolipids; phosphate modifications yield phospholipids; and carbohydrate modifications yield glycolipids. Complex lipids may involve a combination of these characteristics and are named accordingly (ie, glycosphingolipid, glycerophospholipid).

Sphingolipids

Lipids are broadly classified as either hydrolyzable or nonhydrolyzable. They can also be grouped by function as structural lipids, energy-storing lipids, and signaling lipids. Structural lipids are found in the membranes surrounding cells and organelles and include hydrolyzable glycerophospholipids and sphingolipids as well as nonhydrolyzable cholesterol. The properties of individual lipids such as the number of fatty acid tails, double bonds, and head group structures influence both fluidity and curvature of biological membranes. The sphingolipids in Table 3 contain a sphingosine backbone attached to a single fatty acyl chain and a polar head group. In red blood cells and other cell types these sphingolipids are essential structural components of the plasma membrane (Number I). (Number II) The lipids that are primarily responsible for energy storage are triacylglycerides, which are metabolized through hydrolysis and beta-oxidation. The fatty acyl groups of sphingolipids may be hydrolyzed and used for energy production, but this is not their primary function. (Number III) Glycerophospholipids contain two hydrolyzable fatty acyl chains, but sphingolipids contain only one. The other long chain hydrocarbon in a sphingolipid is part of the sphingosine backbone and cannot be hydrolyzed. Therefore, sphingolipids produce one fatty acid upon hydrolysis, not two. Educational objective:Sphingolipids are structural lipids that help influence the fluidity and curvature of biological membranes. The long hydrocarbon chain of the sphingosine head group cannot be readily hydrolyzed, and sphingolipids are not a primary means of energy storage.

hairpin

Nucleic acids base-pair with each other according to Watson-Crick hydrogen bonding rules. RNA is typically single-stranded and may wrap around on itself to form base pairs between nucleotides within the same strand. For example, RNA commonly forms structures known as hairpins, in which a loop with no base pairing is adjacent to a stretch of several base pairs. Structures such as hairpins are stabilized by hydrogen bonds, and the more hydrogen bonds that are present, the more stable the structure. In RNA, base pairing occurs between adenine (A) and uracil (U), or between cytosine (C) and guanine (G). The structures of A and U facilitate the formation of 2 hydrogen bonds in A-U base pairswhereas G-C pairs form 3 hydrogen bonds with each other. Because G-C base pairs contain more hydrogen bonds than A-U base pairs, they are stronger and more stable. For RNA hairpins that have the same number of base pairs, the hairpin that contains the most G-C pairs will be the most stable. For hairpins that have the same number of G-C pairs, the one with the most total base pairs will be the most stable. Choice C shows a hairpin that contains 8 total base pairs, 4 of which are G-C pairs. Accordingly, it is the most stable of the options given. (Choice A) This hairpin has 8 total base pairs, 3 of which are G-C pairs. It has fewer G-C pairs than Choice C, so it is less stable. (Choice B) This hairpin contains 7 total base pairs and 3 G-C pairs. It is the least stable of the options shown. (Choice D) This hairpin has 4 G-C pairs, but only 7 total base pairs. It has fewer total pairs than Choice C, so it is less stable. Educational objective:RNA structures such as hairpin loops are stabilized by hydrogen bonds between base pairs. G-C base pairs form more hydrogen bonds than A-U base pairs. Therefore, structures with more total base pairs and more G-C base pairs are more stable.

pentofuranose

Nucleotide triphosphates (NTPs) consist of a nitrogenous base (adenine, guanine, cytosine, or uracil), a sugar(ribose), and a triphosphate. Ribose is a pentose (five-carbon sugar) that links to the nitrogenous base through a glycosidic bond. A phosphoester bond links the 5′ carbon of ribose to the triphosphate group. Carbohydrates can participate in glycosidic linkages only when in cyclic form (furanose or pyranose), so ribose must be cyclic in NTPs. In the furanose form, the 5′ hydroxyl group can readily participate in phosphoester bonds. Because the ribose in an NTP is a pentose in the furanose form, it is classified as a pentofuranose. (Choice B) In the pyranose form, the 5′ hydroxyl group of ribose becomes part of the ring structure and cannot participate in phosphoester bonds, so ribose cannot be in the pyranose form in NTPs. (Choices C and D) Ribose is a pentose (five carbons), not a hexose (six carbons). Educational objective:Ribonucleotide triphosphates contain a nitrogenous base, ribose, and a triphosphate group linked to the 5′ carbon of ribose. Ribose is a five-carbon sugar (pentose) that must adopt the furanose form to be incorporated into the nucleotide triphosphates. In this form, it can be classified as a pentofuranose.

Purines versus pyrimidines

Nucleotides are classified based on the aromatic compounds from which their nitrogen bases are derived: pyrimidine or purine. The compound pyrimidine is composed of a single aromatic ring containing two nitrogen atoms, which is the basis for thymine (or uracil in RNA) and cytosine. The compound purine is composed of two aromatic rings. It is essentially pyrimidine fused to a five-member imidazole ring, which is the basis for adenine and guanine. The mnemonic "PURe As Gold" is helpful in remembering that purines are associated with A and G nucleotides; the remaining nucleotides are pyrimidines. According to Table 1, the D187V variant of GalC arises due to an A-to-T mutation in the DNA sequence. Adenine is a purine nucleotide, and thymine is a pyrimidine. Therefore, the D187V form of GalC results from replacement of a purine (A) by a pyrimidine (T). (Choice A) Adenine is a purine, not a pyrimidine. (Choice B) Replacement of a pyrimidine by a purine would be correct if the mutation were thymine to adenine instead of adenine to thymine. (Choice D) Thymine is a pyrimidine, not a purine. Educational objective:Nucleotides are classified as purines and pyrimidines based on the molecule from which they are derived. Pyrimidines contain a single aromatic ring (eg, cytosine, thymine, uracil). Purines contain two fused aromatic rings (eg, adenine, guanine).

Phospholipids

Phospholipids (phosphatides) are the main structural lipids of the cell membrane. They are composed of a hydrophilic polar head group, which contains a phosphate, and a hydrophobic tail with one or two fatty acid chains attached to a carbon backbone. Each phospholipid has a unique mass, charge, and solubility due to distinct features in the backbone, polar head groups, and fatty acid chains. They may be separated based on each of the following properties: The charge on phospholipids is determined by the structure of the polar heads. The phosphate component of the head group may be linked to various chemical groups. Positively charged groups such as choline or ethanolamine neutralize the negative charge of the phosphate, whereas neutral groups such as serine, inositol, or glycerol allow the head to maintain a net negative charge. Neutral and negative phospholipids may be separated by high-performance liquid chromatography (Number I). The mass of phospholipids is established by the molecular weight of the backbone and the polar head groups. Phospholipid backbones can contain either a glycerol with two fatty acid tails or sphingosinewith one fatty acid tail. Glycerophospholipids tend to be larger than sphingophospholipids because they have two fatty acid tails instead of one. Polar head groups may be as small as a phosphate (phosphatidic acid) or as large as phosphatidylglycerol. They can be separated by size-exclusion chromatography (Number II). Phospholipid solubility depends on both the hydrophobic ("water-fearing") and hydrophilic ("water-loving") regions of the molecule. Charged head groups are more soluble than neutral groups because they more readily form hydrogen bonds, whereas lipids with long hydrophobic chains have decreased solubility. Phospholipids with different solubilities can be separated by thin-layer chromatography (Number III). Educational objective:Phospholipids are composed of a hydrophilic head group that contains a phosphate and a hydrophobic tail with one or two fatty acid chains. Phospholipids have either a glycerol or a sphingosine backbone. The composition of the head groups and the length of the tails create lipids with distinct solubility, charge, and mass.

Ketone bodies and starvation

Starvation causes a significant decrease in blood glucose levels and stimulates the release of the hormone glucagon, which signals for the synthesis of glucose from glycogen stores (glycogenolysis). After glycogen stores are depleted, the liver begins to break down triacylglycerol reserves. Triacylglycerols are esters of glycerol with three fatty acid chains that function as the main energy reserves in animals. These storage lipids are formed in fat cells, or adipocytes, and yield the most ATP when broken down in the liver into glycerol and free fatty acids. Glycerol is rapidly converted to dihydroxyacetone phosphate for glucose synthesis (gluconeogenesis) while free fatty acids enter the mitochondria, where they are oxidized to generate acetyl-CoA for the citric acid cycle. The accumulated acetyl-CoA is converted into ketone bodies such as acetoacetate and 3-hydroxybutyrate. Although the brain primarily uses glucose as a fuel source, it can also use ketone bodies during times of starvation. Based on the passage, patients with hypertriglyceridemia have a surplus of triacylglycerol in their bloodstream. During starvation, excess triacylglycerol in the bloodstream and liver will ultimately be converted into ketone bodies in these patients, and will be utilized by the brain and other tissues. (Choice A) Glycerol is rapidly converted into glycerol 3-phosphate and then oxidized into dihydroxyacetone phosphate during prolonged starvation. It does not accumulate. (Choice C) Glycogen stores are depleted within 12-24 hours of fasting and are not available after prolonged starvation. (Choice D) Prolonged starvation causes amino acids in proteins to be broken down to synthesize glucose. Educational objective:Triacylglycerols, also known as triglycerides, are storage lipids that consist of three fatty acid chains attached to a glycerol molecule. During prolonged starvation, acetyl-CoA produced from fatty acid oxidation in the mitochondria is converted into alternative fuel molecules known as ketone bodies.

Steroid hormones synthesized from cholesterol

Steroid hormones are involved in multiple critical signaling pathways in the body, including those related to physical development, reproduction, and stress. All steroid hormones are synthesized from cholesterol and share a common backbone structure consisting of four fused rings: 3 six-member rings and 1 five-member ring. The basic unit of cholesterol is a branched five-carbon molecule belonging to a class known as isoprenes. Isoprenes can combine with each other to form larger units known as terpenes. Terpenes consisting of two isoprene units are called monoterpenes, those consisting of two monoterpenes (or four isoprenes) are called diterpenes, and those consisting of three monoterpenes (six isoprenes) are called triterpenes. During cholesterol synthesis, a triterpene known as squalene forms. Squalene undergoes multiple steps, including demethylation, methyl rearrangements, and cyclization, to form cholesterol. The correct order of synthesis for the precursors of steroid hormones is isoprene → monoterpene → squalene → cholesterol. (Choices B, C, and D) Each of these options either has something other than cholesterol as the final product or something other than isoprene as the starting material. Steroid hormones are synthesized from cholesterol, which is made of several isoprene units linked together. Educational objective:Steroid hormones are derived from cholesterol. Cholesterol has a characteristic four-ring backbone, which is synthesized from five-carbon subunits called isoprenes. Two isoprenes joined together form a monoterpene, and six isoprenes can join to form a triterpene called squalene, which then cyclizes and, after several steps, forms cholesterol.

Statin therapy

Steroids are tetracyclic derivatives of terpenes that contain four hydrocarbon rings fused together (three 6-membered rings and one 5-membered ring. Cholesterol is a steroid alcohol and the precursor molecule for the synthesis of bile salts, vitamin D, and steroid hormones. Cholesterol and its hormone derivatives are classified as amphipathic molecules because they contain hydrophobic and hydrophilic regions, which allow them to diffuse across cell membranes. Common steroid hormones include testosterone, cortisol, aldosterone, estrogen, and progesterone, all of which are secreted by endocrine glands into the bloodstream, where they travel to their target cells. According to the passage, statin therapy treats atherosclerosis by inhibiting the production of a cholesterol precursor. Therefore, statins will decrease the levels of cholesterol and consequently the levels of steroid hormones such as testosterone. (Choices A and C) Peptide hormones such as leptin and insulin are made from chains of amino acids. Neither would be affected by the reductions in cholesterol synthesis caused by statin therapy. (Choice D) Cortisol concentration would be reduced in individuals undergoing statin therapy because cholesterol is necessary for cortisol production. Educational objective:Cholesterol is a steroid alcohol and the precursor molecule for the synthesis of steroid hormones, bile salts, and vitamin D. Common steroid hormones include cortisol, aldosterone, estrogen, testosterone, and progesterone. Inhibition of cholesterol synthesis reduces the levels of steroid hormones.

Terpenes

Terpenes are precursors in the synthesis of cholesterol and steroid hormones. They are derived from isoprenes and do not play a role in inflammatory responses.

GC, AT Hydrogen bonds and melting temperatures

The GC content is the percentage of guanine and cytosine bases in DNA, and the higher the GC content, the greater the stability and melting temperature (Tm) of double-stranded DNA. DNA double helix stability is primarily maintained by stacking interactions: the interactions between neighboring G≡C base pairs are stronger than those between adjacent A=T base pairs. It also takes more energy (heat) to break the three hydrogen bonds in G≡C base pairs than the two hydrogen bonds in A=T base pairs. As a result, the Tm for GC-rich DNA molecules is higher as additional energy is required to separate the more stable helix. Figure 1 shows that SW-480 cells have a G→T base substitution at codon 12 of the K-ras gene, depicting a loss in GC content. Assuming there are additional similar mutations in the mutant K-ras gene, the mutant DNA will have lower GC content than the WT DNA and therefore be less stable and denature faster (ie, exhibit lower Tm). (Choice A) Mutant K-ras DNA with various mutations similar to the one shown in Figure 1 would exhibit a loss of GC content and decreased stability and Tm. (Choices C and D) WT DNA and mutant K-ras DNA are the same length because the mutation in K-ras from SW-480 cells is a base substitution and not an insertion or deletion. Longer DNA molecules have increased Tm because they have more base pairs that stabilize the double helix and require more energy to be broken. Educational objective:G≡C base pairs form more hydrogen bonds and have stronger stacking interactions than A=T base pairs. The higher the GC content, the greater the melting temperature (Tm) and stability of double-stranded DNA.

Michaelis-Menten constant

The Michaelis-Menten constant Km is the substrate concentration at which half of the reaction's maximum velocity is achieved. It depends on both the rate of substrate binding to the enzyme (ES complex formation) and the rate at which the bound substrate is converted to product (product formation). If the rate of product formation is significantly slower than the rate of ES complex formation, as indicated in the question, then Km essentially measures an enzyme's affinity for a substrate, or the tendency of an enzyme and a substrate to bind and form a complex. When product formation is rate-limiting, Km can also be thought of as the substrate concentration at which half of the enzymes in solution are bound. A small Km indicates a strong ES complex because only a small amount of substrate is required to achieve a large amount of complex formation. On the other hand, a large Km indicates a weak ES complex that does not readily form. According to Table 1, CSL-174 has a smaller Km for Hip1 than WKLL-ACC does, so CSL-174 forms a stronger ES complex. (Choice A) CSL-175 has a smaller Km than WKLL-ACC, which means that WKLL-ACC forms a weaker ES complex. In fact, WKLL-ACC has the highest Km of all, and therefore forms the weakest complex. (Choice C) CSL-176 has the smallest Km of the peptides tested, and therefore forms the strongest ES complex. (Choice D) CSL-175 has a smaller Km than CSL-174, so it forms a stronger complex. Educational objective:The Michaelis-Menten constant Km is an approximate measure of an enzyme's affinity for a substrate or of the strength of the enzyme-substrate complex. A low Km represents a high affinity and a strong complex whereas a high Km indicates a low affinity and a weak complex.

Fatty acid structure and membrane fluidity

The fluidity and rigidity of plasma membranes is partially dependent on the structure of fatty acyl tails in the lipid bilayers. Increasing hydrophobic interactions between fatty acyl tails leads to decreased motion and more rigidity. The length and number of double bonds in a fatty acyl tail influence the number of hydrophobic interactions possible. Short tails participate in fewer hydrophobic interactions than long tails and therefore result in increased fluidity. Carbon-carbon double bonds in fatty acyl tails are typically in the cis-conformation and introduce a bend in the tail, preventing adjacent tails from interacting efficiently with each other. A higher number of double bonds leads to decreased interaction and greater fluidity. The lipid in Table 3 that has the shortest fatty acyl tail length and the most double bonds will make the greatest contribution to membrane fluidity. GL6 and GL3 have the shortest fatty acyl tails (20 carbons), but GL6 has more double bonds (one instead of zero). Therefore, GL6 makes the greatest contribution to membrane fluidity. (Choice A) GL2 has the longest tail of the lipids in the table and has zero C=C double bonds. It makes the smallest contribution to membrane fluidity. (Choice B) GL3 and GL6 both have the shortest tail lengths, but GL3 has no C=C double bonds, so it makes a smaller contribution to membrane fluidity than GL6. (Choice C) GL5 has a double bond, but it has a longer tail than GL6, so GL5 makes a smaller contribution to membrane fluidity. Educational objective:The fluidity of a cell membrane is largely dependent on the lengths of the fatty acyl tails in the bilayer and on the number of double bonds in each tail. Short chains with double bonds yield the highest fluidity because they participate in the fewest hydrophobic interactions with neighboring lipids.

A, T, C, G look

The four nucleotides found in DNA (adenine, cytosine, guanine, and thymine) can be identified by their ring structures and by the number of Watson-Crick hydrogen bond donors and acceptors on each. The purines(adenine and guanine) each have two rings whereas the pyrimidines (cytosine and thymine) each have one ring. Hydrogen bond donors can be either a nitrogen or an oxygen atom with hydrogen attached. Acceptors can take the form of oxygen or nitrogen with a lone electron pair. Guanine and cytosine each have three hydrogen bond participants: An acceptor and two donors in guanine, and a donor and two acceptors in cytosine. Adenine and thymine have one donor and one acceptor each involved in base pairing. Figure 1 shows that the B-antigen is produced by galactosyl transferase. Table 1 shows that the galactosyl transferase allele has adenine (A) at position 793. Adenine is represented in Choice D, which shows a purine with one hydrogen bond donor and one acceptor. (Choice A) This option shows thymine, which is not present at position 793 in either allele of the ABO gene. (Choice B) This option shows guanine which, like adenine, is a purine. Unlike adenine, guanine has one hydrogen bond acceptor and two donors. (Choice C) This option shows cytosine, which is the nucleotide at position 793 for the GalNAcyl transferase allele. This allele corresponds to type A blood (or type AB), but not type B blood. Educational objective:Nucleotide bases can be identified by their ring structure (purine or pyrimidine) and by the number of Watson-Crick hydrogen bond donors and acceptors. Adenine and guanine are purines, whereas thymine and cytosine are pyrimidines. Thymine and adenine have one donor and one acceptor each, whereas guanine has an acceptor and two donors and cytosine has two acceptors and one donor.

Fat soluble vitamins

The human body is capable of using precursor molecules to synthesize many of the biomolecules needed for proper function. However, humans lack the enzymes necessary to synthesize certain molecules such as the essential amino and fatty acids, and the class of molecules known as vitamins. These molecules must be obtained from the diet. Vitamins are classified as either water-soluble (B series, C) or fat-soluble (A, D, E, K). Water-soluble vitamins contain several hydrophilic groups and are easily dissolved in aqueous environments. They are not stored in the body, and the excess is excreted in the urine (Choice B). For this reason, water-soluble vitamins must be frequently replenished. In contrast, fat-soluble vitamins are highly hydrophobic. They dissolve in lipid rather than aqueous environments and cannot be easily excreted in urine. Instead, the excess is stored and accumulates in adipose tissue, which consists mainly of fat cells. (Choice A) Steroid hormones are produced primarily in the adrenal cortex and the gonads from cholesterol and other precursors. They are not required in the diet and do not generally accumulate in adipose tissue. (Choice C) Peptide hormones are readily synthesized by the body, and are not required in the diet. They are generally soluble in water, and are not stored in adipose tissue. Educational objective:Certain biomolecules such as vitamins and the essential amino and fatty acids cannot be synthesized in the human body and must be obtained through the diet. Water-soluble vitamins (B series, C) are excreted in the urine whereas fat-soluble vitamins (A, D, E, and K) are stored in adipose and other fatty tissues.

Effect of hydrogen bonding on DNA melting temperature

The melting temperature Tm of DNA is the temperature at which 50% of the double helices in solution have been separated into single strands. It is determined by the strength of the intermolecular forces holding the strands together. An increased number of intermolecular interactions increases the Tm because more energy is required to disrupt them. Hydrogen bonding between bases (Watson-Crick hydrogen bonding) is one of the major intermolecular forces that hold DNA strands together. The number of hydrogen bonds between base pairs differs: G and C share three hydrogen bonds whereas A and T share only two. Therefore, DNA molecules with high GC content have higher melting temperatures than those with low GC content. The passage states that dFTP lacks Watson-Crick hydrogen bonding. Therefore, dFTP incorporation in place of dTTP will weaken interactions with adenine. Because there are fewer hydrogen bonds holding the strands together, less energy will be required to separate them. Consequently, DNA that incorporated dFTP instead of dTTP will most likely have a decreased melting temperature. (Choice A) The melting temperature is expected to decrease. Reduced base stacking would most likely decrease, not increase, Tm due to decreased hydrophobic interactions. (Choice C) Although dFTP is similar in shape to dTTP and specifically pairs with adenine, its inability to form hydrogen bonds with adenine makes the interaction weaker and results in a decreased melting temperature. (Choice D) The melting temperature of a DNA molecule is affected by the number of nucleotides. DNA molecules with more nucleotides have more intermolecular interactions (eg, hydrogen bonds) and increased melting points. However, the question states that the DNA molecule has a fixed length, so the number of nucleotides is not a factor in this scenario. Educational objective:The strands of a DNA double helix are held together by intermolecular interactions that include hydrogen bonds. Reducing the number of hydrogen bonds decreases the melting point of a double helix.

Target species of influenza

The passage suggests that the G225D mutation allows influenza A virions to switch their target species from birds to humans. An experiment designed to determine whether H6-G225D virions exclusively infect humans should include controls that are known to infect either human or avian species only. The effects of these control groups on cells can then be compared to the effects of H6-G225 in each type of cell. A positive control displays the response that would be expected if H6-G225D is able to infect human cells. Because H1N1 is known to infect human tissue exclusively, it works as an adequate positive control (Number II); if the G225D mutation allows virions to infect human cells, then the effects of H1N1 and H6-G225D on human tracheal cells will be similar. A negative control shows the response that would be expected if the virus did not switch target species: infection of avian cells but not human cells. WT-H6N1 serves as a negative control because it exclusively infects avian tracheal tissue (Number I). If the G225D mutation eliminates the affinity of virions for avian tissue, H6-G225D will not show a response in avian cells whereas WT-H6N1 will. In a well-designed experiment, both positive and negative controls are needed to validate the researcher's predictions and results. (Number III) The double mutant does not cause infection in either of the two target species of influenza A. It is not necessary as a control because it has the same effect as no virus at all. Educational objective:Controls are necessary experimental tools that are used to compare and validate the results of an experiment. Positive controls display an expected response; in negative controls, no response is expected.