Question 1

Passage The K-ras gene codes for a small guanine nucleotide-binding protein, which acts as a molecular activator in numerous signaling pathways that promote DNA replication along with cell growth and survival. K-ras mutations at codons 12 and 13 function as biological markers for diagnosis of tumors. The DNA of SW-480 cells, a line originating from a colon adenocarcinoma, is the standard positive control used to identify codon 12 mutations in K-ras testing. Figure 1 Structural comparison of K-ras codon 12 in wild-type cells and SW-480 cellsPCR sequence analysis has been the preferred method for studying genetic mutations such as those in K-ras. However, challenges in analyzing PCR amplicons have prompted the development of DNA melting curve analysis (MCA). In the MCA technique, DNA is extracted from clinical samples and amplified by PCR. The amplified DNA is then melted (30-70°C) and analyzed by assessing the fluorescence (at 640 nm) of LC Green Plus, an intercalating dye that fluoresces when bound to double-stranded DNA (Figure 2). Figure 2 Melting curves of wild-type (WT) and SW-480 K-ras DNA.To study K-ras activity in colorectal cancer cells, short tandem repeats (STRs) were introduced into the K-ras gene. STRs are short stretches of DNA with repeating nucleotides, and four different STR sequences were inserted into the K-ras gene and analyzed via MCA under different experimental conditions (Table 1).Table 1 Sequences Studied in K-ras DNA MCA Analysis. Liu YP, Wu HY, Yang X, et al. Diagnostic accuracy of high resolution melting analysis for detection of KRAS mutations: a systematic review and meta-analysis. Sci Rep. 2014;4:7521. -Prior to inserting the STR sequences into K-ras DNA, the researchers perform MCA analysis using only the double-stranded STR sequences. Which sequence from Table 1 would have the fastest reannealing time in this experiment?

Accepted Answer

11ecba2d_0e71_17d9_a3bf_816fbff27825_MD0008_00 Melted DNA strands will reanneal to form the double helix once temperature decreases, but the time it takes for complementary strands to reanneal depends on the following: DNA length: Longer DNA molecules have more hydrogen bonds and will take more time to both melt and reanneal. pH: The physiological pH range (7.3-7.4) allows maximal hydrogen bonding between nitrogenous bases of DNA. At low pH, hydrogen bond acceptor atoms in the bases become protonated, and these protonated acceptors cannot form hydrogen bonds, causing double helix separation. In contrast, high pH causes deprotonation of hydrogen bond donors in the bases; the loss of protons results in the loss of hydrogen bonds and a destabilized double helix. Salt concentration (ionic strength): The electrostatic repulsion between negatively charged phosphate groups on the sugar-phosphate backbone destabilizes the double helix. However, this repulsion is neutralized and shielded by the binding of positively charged species in solution (eg, Na⁺ and Mg²⁺ cations). High salt concentration of the solution increases double helix stability, but low salt concentration decreases stability. 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.

Question 2

A solution of which amino acid will NOT rotate plane polarized light?

Accepted Answer

11ecba2d_0e67_53cd_a3bf_f7c66bdfd16f_MD0008_00 Molecules with at least one atom bonded to four different chemical groups (a stereocenter) are said to be chiral. The bonds to any stereocenter can be arranged in two distinct ways, yielding two possible versions of each stereocenter in a chiral molecule. Solutions of chiral molecules rotate plane polarized light, and chiral molecules with opposite arrangements at every stereocenter (enantiomers) rotate light by the same amount but in opposite directions. Amino acids contain an amino group and a carboxylate group linked together through an alkyl carbon, known as the alpha carbon. The alpha carbon is also bonded to a hydrogen atom and to one other chemical group, called the R-group or the side chain. If the side chain is different than the other three substituents, the amino acid is chiral and is given an L- or D- designation. L-amino acids predominate in nature. In biological systems, the side chain can be one of 20 chemical groups associated with the standard amino acids. The side chain of glycine is a hydrogen atom, so the alpha carbon of glycine is bonded to two hydrogen atoms. Because the two hydrogen atoms are identical, the glycine alpha carbon is not bonded to four distinct chemical groups. Therefore, glycine is achiral (neither L nor D) and will not rotate plane polarized light. (Choices A, C, and D) Apart from glycine, all of the standard amino acids have alpha carbons bonded to four unique substituents. Therefore, they are chiral and will rotate plane polarized light. Educational objective: Molecules with one or more stereocenters (chiral molecules) rotate plane polarized light. A stereocenter forms when one atom is bonded to four unique groups, and can be arranged in either of two ways. All of the standard amino acids are chiral except glycine.

Question 3

Passage Nerve cells must maintain the composition of their myelin sheaths to function properly. Galactocerebroside (GC), shown in Figure 1, is a hydrolyzable lipid and an important component of the myelin sheath. Figure 1 Structure of the myelin sheath component galactocerebrosideHomeostasis of GC is maintained by its dynamic synthesis and degradation, as shown in Figure 2. GC is synthesized by the enzyme ceramide galactosyltransferase (CGT) using UDP-galactose (UDP-Gal) and ceramide (Cer) as substrates. It is degraded by the enzyme galactocerebrosidase (GalC), which hydrolyzes the glycosidic bond to release galactose (Gal) and ceramide. A deficiency in GalC activity leads to a buildup of GC, and prior research has connected this buildup with neurodegeneration. Figure 2 Schematic of GCT and GalC activity in healthy and GalC deficient cellsOver 100 inactivating GalC mutations have been identified. Five of the most common mutations are listed in Table 1. The changes to both DNA and amino acid sequence are shown.Table 1 Five Common Galactocerebrosidase Mutations -GC can be classified as:

Accepted Answer

11ecba2d_0e66_425c_a3bf_31f73af1913f_MD0008_00 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).

Question 4

Proline differs from the other standard amino acids in that only the former:

Accepted Answer

11ecba2d_0e68_171e_a3bf_199d69021b67_MD0008_00 The core structure of amino acids, called the backbone, consists of an amino group linked to an alkyl carbon (the α-carbon), which is linked to a carboxyl group. The α-carbon is also linked to an R group, which varies from one amino acid to another and determines the identity of each amino acid. Twenty standard amino acids, each with a unique R group, are found in proteins in all domains of life. Proline is unique among the standard amino acids because its R group forms a ring that includes the backbone amino group. Because of this, the amino group of the proline backbone has two bonds to carbon, making it a secondary amine. In contrast, all other standard amino acids have primary amines in their backbones. (Choice A) The acidic amino acids (aspartate and glutamate) have negatively charged side chains. Proline does not. (Choice B) Peptide bonds are amide bonds. Due to resonance, all amide bonds (including peptide bonds) have partial double-bond character, and therefore all are planar, and not just the bond in proline. (Choice D) The R group of proline is bonded to the backbone at two points instead of one, making it relatively rigid and unlikely to be found in the flexible regions of a protein. Note that β-turns, which commonly contain proline, are highly structured and inflexible. In addition, many amino acids are found in flexible-loop regions, so even if it were common, proline's inclusion in flexible loops would not be a unique property of proline. Educational objective: All amino acids have a backbone consisting of an amino group, an α-carbon, and a carboxyl group. Each amino acid also has an R group that determines its identity. Proline is unique among the standard amino acids because its R group bonds to the backbone amino group, making it a secondary rather than a primary amine.

Question 5

If a stretch of double-stranded DNA has a total of 560 nucleotides and 159 adenine nucleotides, what is the nucleotide composition of this DNA molecule?  It has:

Accepted Answer

11ecba2d_0e73_61cb_a3bf_1f571a59fe98_MD0008_00 Purines (guanine or adenine) base pair with pyrimidines (cytosine or thymine) in DNA. Adenine (A) base pairs exclusively with thymine (T) via two hydrogen bonds, and guanine (G) base pairs with cytosine (C) via three hydrogen bonds. As a result, Chargaff's rules state that there is a 1:1 nucleotide mole ratio of pyrimidines to purines and that for any double-stranded DNA molecule: The amount of guanine equals the amount of cytosine, and the amount of adenine equals the amount of thymine (ie, G=C and A=T) The sum of purines equals the sum of pyrimidines (ie, G+A = C+T) If a stretch of DNA has 560 total nucleotides and 159 of these are adenine nucleotides, then there are also 159 thymine nucleotides (Choices A, B, and D). Nucleotide composition can be calculated as follows: Obtain the number of guanine and cytosine nucleotides by subtracting the number of adenine (159) and thymine (159) nucleotides from the total number of nucleotides (560). 560 - (2 x 159) = 560 - 318 = 242 nucleotides Divide 242 by 2 to obtain the number of guanine and cytosine nucleotides. 242/2 = 121 There are 121 guanine and 121 cytosine nucleotides in the described DNA. Educational objective: Chargaff's rules state that there is a 1:1 nucleotide mole ratio of pyrimidines to purines and that for any double-stranded DNA sample: A=T and G=C G+A = C+T

Question 6

Which factor decreases the stability of a DNA double helix during denaturation?

Accepted Answer

11ecba2d_0e72_776a_a3bf_8dbc2a18919e_MD0008_00 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.

Question 7

Which of the following molecules could decrease the activity of one or more Krebs cycle enzymes?

Accepted Answer

11ecba2d_0e6a_12f1_a3bf_0130f28cd833_MD0008_00 The Krebs cycle, also known as the citric acid cycle, is tightly regulated to produce ATP at the optimal rate for the needs of the cell. The cycle includes three irreversible steps, catalyzed by the enzymes citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase. The activity of each of these enzymes can be increased or decreased when specific molecules bind to the enzymes and alter their shape (allostery). Each of these molecules serves as an indicator of the energy state of the cell. Molecules that are associated with depleted energy (eg, ADP) and muscle contractions (eg, calcium) indicate a need for additional ATP production. These molecules allosterically activate Krebs cycle enzymes to meet the demand for energy. Conversely, molecules that indicate that the Krebs cycle is producing sufficient energy allosterically inhibit the enzymes. These molecules include several products of the Krebs cycle and the electron transport chain such as ATP, citrate, succinyl-CoA, and NADH. NADH decreases the activity of all three of the regulated Krebs cycle enzymes. (Choice A) Glucagon is a hormone that upregulates glucose production in the liver through glycogen breakdown (glycogenolysis) and gluconeogenesis. The enzymes of the citric acid cycle are not under hormonal control. (Choice C) Acetyl-CoA reacts with oxaloacetate to form citrate at the beginning of the Krebs cycle. Its presence increases the overall rate of the cycle by increasing substrate concentration. (Choice D) ADP is produced when ATP is hydrolyzed, and it signals to the cell that energy has been depleted. It allosterically activates the Krebs cycle enzymes citrate synthase and isocitrate dehydrogenase, thereby increasing their activity. Educational objective: The Krebs cycle is allosterically regulated at the irreversible steps catalyzed by citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase. The enzymes are inhibited by NADH, ATP, citrate, and succinyl-CoA, and are activated by ADP and calcium. The Krebs cycle enzymes are not under hormonal control.

Question 8

The compound 5,10-methylene tetrahydrofolate (5,10mTHF) is derived from folic acid (vitamin B₉) and is required in the active site of the enzyme thymidylate synthase for thymine synthesis. Therefore, 5,10mTHF is most likely:

Accepted Answer

11ecba2d_0e75_d2cf_a3bf_f7a5edd444f9_MD0008_00 Enzymes are catalysts that facilitate reactions in biological settings. They consist predominantly of amino acids, but many enzymes require non-amino acid components called cofactors (inorganic) and coenzymes (organic) to function. Many coenzymes, including THF, NADH, FADH₂, and coenzyme A, are derived from vital organic compounds that can be obtained only through the diet, collectively called vitamins. The question states that 5,10-methylene tetrahydrofolate (5,10mTHF) is derived from a vitamin (B₉) and that it is required for enzymatic function of thymidylate synthase. Therefore, 5,10mTHF is most likely acting as a coenzyme. (Choice A) Transcription factors bind to DNA and alter transcription levels by affecting RNA polymerase binding. They are not involved in the synthesis of individual nucleotides such as thymine. (Choice B) Inhibitors reduce the activity of an enzyme. According to the question, 5,10mTHF is required for enzymatic function. (Choice C) Allosteric activators increase the activity of an enzyme by binding somewhere other than the active site and changing the shape of the enzyme. The question states that 5,10mTHF is required in the active site, so it does not act allosterically. Educational objective: Enzymes frequently require non-amino acid components to carry out their functions. These components are referred to as cofactors and coenzymes, and many of them are derived from vital organic compounds called vitamins.

Question 9

The pK_a for the side chain of cysteine is 8.4. What is the ratio of protonated side chain (-SH) to deprotonated side chain (-S⁻) in a solution of cysteine at physiological pH of 7.4?

Accepted Answer

11ecba2d_0e78_e016_a3bf_4b05dcae3485_MD0008_00 Some chemical groups, including the side chains of some amino acids, can be protonated or deprotonated depending on the pH of the surrounding environment. The pK_a of a chemical group is the pH at which the ratio of protonated to deprotonated species is 1:1. A decrease in pH yields increased H⁺ concentration (favors protonation) whereas an increase in pH yields decreased H⁺ concentration (favors deprotonation). The protonated:deprotonated ratio of a substance can be determined by solving the Henderson-Hasselbalch equation. However, when the pH differs from the pK_a by an integer number of pH units, the ratio can be determined more easily because each pH unit alters the ratio by a factor of 10. The question states that the pK_a of the cysteine side chain is 8.4 and the solution is at a pH of 7.4. Because the pH of the solution is lower than the pK_a of cysteine, the protonated form of cysteine will be favored. Furthermore, because the pH is exactly 1 pH unit lower than the pK_a, protonated cysteine will exceed the deprotonated form by a factor of 10. Therefore, the ratio of protonated to deprotonated cysteine side chain at physiological pH is 10:1. (Choice A) 1:100 is the ratio of protonated cysteine side chain to deprotonated side chain at a pH of 10.4. (Choice B) 1:10 is the ratio of protonated cysteine side chain to deprotonated side chain at a pH of 9.4. It is also the deprotonated:protonated ratio at physiological pH (reciprocal of the correct ratio). (Choice D) 100:1 is the ratio of protonated cysteine side chain to deprotonated side chain at a pH of 6.4. Educational objective: The pK_a of a chemical group is the pH at which the ratio of protonated to deprotonated species is 1:1. Decreasing the pH below the pK_a yields a higher proportion of protonated species, and increasing the pH yields a lower proportion. Each pH unit above or below the pK_a alters the ratio of protonated to deprotonated species by a factor of 10.

Question 10

A fatty acid undergoes complete β-oxidation to yield nine molecules of acetyl-CoA.  No isomerization reactions occur.  The structure of the original fatty acid was:

Accepted Answer

11ecba2d_0e77_0b54_a3bf_cd806aee5e9b_MD0008_00 Fatty acids are long hydrocarbon chains with a carboxylic acid group on one end. They are degraded through a process called fatty acid oxidation, or β-oxidation, which proceeds differently depending on the composition of the fatty acid being oxidized. All fatty acids are broken down into multiple two-carbon units called acetyl-CoA. Fatty acids with an even number of carbons yield half as many acetyl-CoA units as the total number of carbons in the original molecule. If the fatty acid has an odd number of carbons, the final three carbons are converted into a compound called propionyl-CoA. In other words, for odd-chain fatty acids, the number of acetyl-CoA units produced equals (total carbons − 3)/2. Unsaturated fatty acids have at least one C=C double bond, typically in the cis-conformation. For β-oxidation to proceed, the cis-bonds must be isomerized to trans-bonds, which are introduced as part of the oxidation mechanism. Saturated fatty acids do not have C=C double bonds initially and do not need to undergo isomerization during β-oxidation. The question states that the β-oxidation resulted in 9 acetyl-CoA units, so the original fatty acid must have had twice that number of carbons (18). The fact that no isomerization reactions were needed indicates that the fatty acid was saturated. Only Choice B shows a saturated fatty acid with 18 carbons. (Choice A) This option shows a saturated fatty acid with 19 carbons. Because it has an odd number of carbons, it will produce 8 acetyl-CoA units (from 16 carbons) and have 1 three-carbon unit (called propionyl-CoA) remaining. (Choices C and D) These options show unsaturated fatty acids, which require isomerization of the cis-double bond. Educational objective: Fatty acids can be oxidized to form acetyl-CoA, and those with an even number of carbons produce half as many acetyl-CoA molecules as they have carbons. Fatty acids with an odd number of carbons produce propionyl-CoA in addition to acetyl-CoA. Unsaturated fatty acids require isomerization reactions to convert cis-bonds to trans-bonds whereas saturated fatty acids do not.

Question 11

An infant who experiences seizures and has lactic acid buildup is shown to have deficient pyruvate dehydrogenase complex (PDHC) activity.  Genetic analysis shows that both the maternal and paternal alleles for all PDHC subunits are normal.  These results could indicate:

Accepted Answer

11ecba2d_0e74_e86e_a3bf_631d1d1ff9f4_MD0008_00 The pyruvate dehydrogenase complex (PDHC) is an enzyme composed of three subunits (E1, E2, and E3). It catalyzes the oxidative decarboxylation of pyruvate to form acetyl-CoA, along with the reduction of NAD⁺ to form NADH. During this process, electrons are passed from one subunit in the complex to the next until they can be transferred to NAD⁺. This transfer is facilitated in part by the cofactor lipoic acid. Cofactors are non-amino acid groups of proteins that are necessary for the protein to function. Because they are not amino acids, cofactors are not directly encoded by DNA. The question states that PDHC has decreased activity, yet the alleles (DNA) encoding it are normal. Therefore, some component of PDHC activity that is not directly encoded in the DNA must be altered. Because lipoic acid is a necessary cofactor of the PDHC, impaired lipoic acid production would result in decreased activity of the PDHC complex. (Choice A) The question states that alleles for all PDHC subunits are normal. Underexpression, not overexpression, could account for decreased activity of the normal proteins. (Choice C) The PDHC synthesizes acetyl-CoA, which enters the Krebs cycle (also known as the citric acid cycle). A deficient PDHC would decrease acetyl-CoA concentration and downregulate its entry into the Krebs cycle. (Choice D) Lactic acid is exclusively produced by pyruvate fermentation, so the fact that the infant experienced lactic acid buildup indicates that the rate of fermentation is high. Educational objective: Many enzymes rely on non-amino acid components called cofactors or coenzymes to function. The pyruvate dehydrogenase complex requires lipoic acid for activity.

Question 12

Under physiological conditions, peptide bond formation and degradation both require enzymes, but only formation requires coupling to GTP hydrolysis.  Based on this information, peptide bonds under physiological conditions are:

Accepted Answer

11ecba2d_0e6b_2462_a3bf_8197cfdc0a8f_MD0008_00 Chemical reactions, including those in biological settings, are controlled both by thermodynamics and kinetics. Thermodynamics describes the energy difference between products and reactants (ΔG), and dictates whether a reaction requires energy input. Kinetics describes the energy required to initiate a reaction (activation energy), and determines how quickly a reaction proceeds. If a reaction has a positive ΔG (as in anabolic processes such as protein synthesis), the product is thermodynamically unstable and the reaction requires energy input. In biological settings, this energy input is usually derived from the hydrolysis of ATP or GTP. The reverse reaction is thermodynamically favorable (negative ΔG) and requires no energy input; however, this does not mean the reverse reaction will proceed quickly. Reactions that are thermodynamically favorable but have high activation energies are slow, and the bonds that are broken in these reactions are said to be kinetically stable. The question states that peptide bond formation requires coupling to GTP hydrolysis, a source of energy. Therefore, formation requires energy input and degradation does not, indicating that peptide bonds are thermodynamically unstable (ie, prone to breaking). However, the question also states that degradation requires enzymes. Enzymes increase the rate of otherwise slow reactions by decreasing the activation energy, so in the absence of enzymes peptide bonds degrade slowly, making them kinetically stable. (Choices A and C) Because their formation requires coupling to an energy source (GTP), peptide bonds are thermodynamically unstable. (Choice D) Although it is true that peptide bonds are thermodynamically unstable, the fact that enzymes are required to break peptide bonds means they are kinetically stable. In the absence of the relevant enzymes (called proteases), peptide bond degradation is so slow as to be undetectable on a physiological time scale. Educational objective: Thermodynamics determines whether a reaction can proceed without energy input, and kinetics determines how quickly a reaction proceeds. A reaction that is thermodynamically favorable but slow involves bonds that are thermodynamically unstable (prone to breaking) but kinetically stable (break slowly). Anabolic processes such as DNA and protein synthesis require energy input (usually from ATP or GTP) but form kinetically stable bonds.

Question 13

Which series depicts the order in which the precursors of steroid hormones are synthesized?

Accepted Answer

11ecba2d_0e77_f5b5_a3bf_299f9121a328_MD0008_00 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.

Question 14

Based on the titration curve of an unknown amino acid, shown below, can the side chain of the amino acid readily exchange protons with water?

Accepted Answer

11ecba2d_0e69_2890_a3bf_3d1722ba5610_MD0008_00 Acidic and basic chemical groups, also known as ionizable groups, have intrinsic tendencies to exchange protons with water. This tendency is described by the pK_a, or the pH at which half of the ionizable groups of interest are protonated. pK_a values can be determined by titration, which is typically performed by gradually adding a base to an acidic solution of the molecule of interest. As the pH of the solution approaches the pK_a of a chemical group, the system acts as a buffer and resists changes in pH by accepting protons from water or donating protons to it. On a titration curve, the region near a pK_a (called the buffering region) is relatively flat because the addition of base changes the pH by only a small amount. All amino acids have ionizable carboxyl and amino groups with pK_a values near 2 and 9.5, respectively. Therefore, all amino acid titration curves have two buffering regions, one near each pK_a. Seven of the standard amino acids (R, K, Y, C, H, E, and D) also have ionizable side chains, each with a unique pK_a. The titration curve for each of these amino acids has a third buffering region associated with the side chain pK_a. Because the titration curve in the question shows only two buffering regions, the amino acid has only two pK_a values. Therefore, its side chain is not ionizable and cannot exchange protons with water. (Choices A and B) The titration curve corresponds to an amino acid with a non-ionizable side chain, which cannot exchange protons with water. It has only two equivalence points and two buffering regions. (Choice C) The final pH of a titration is determined by the pH of the base (or acid) that is gradually added during the titration. It is not related to the molecule that is titrated. Educational objective: All amino acids have at least two pK_a values that correspond to their carboxyl and amino groups. Amino acids with ionizable side groups (R, K, Y, C, H, E, and D) have three pK_a values. Titration curves show the region surrounding each pK_a as relatively flat portions of the curve.

Question 15

Passage The K-ras gene codes for a small guanine nucleotide-binding protein, which acts as a molecular activator in numerous signaling pathways that promote DNA replication along with cell growth and survival. K-ras mutations at codons 12 and 13 function as biological markers for diagnosis of tumors. The DNA of SW-480 cells, a line originating from a colon adenocarcinoma, is the standard positive control used to identify codon 12 mutations in K-ras testing. Figure 1 Structural comparison of K-ras codon 12 in wild-type cells and SW-480 cellsPCR sequence analysis has been the preferred method for studying genetic mutations such as those in K-ras. However, challenges in analyzing PCR amplicons have prompted the development of DNA melting curve analysis (MCA). In the MCA technique, DNA is extracted from clinical samples and amplified by PCR. The amplified DNA is then melted (30-70°C) and analyzed by assessing the fluorescence (at 640 nm) of LC Green Plus, an intercalating dye that fluoresces when bound to double-stranded DNA (Figure 2). Figure 2 Melting curves of wild-type (WT) and SW-480 K-ras DNA.To study K-ras activity in colorectal cancer cells, short tandem repeats (STRs) were introduced into the K-ras gene. STRs are short stretches of DNA with repeating nucleotides, and four different STR sequences were inserted into the K-ras gene and analyzed via MCA under different experimental conditions (Table 1).Table 1 Sequences Studied in K-ras DNA MCA Analysis. Liu YP, Wu HY, Yang X, et al. Diagnostic accuracy of high resolution melting analysis for detection of KRAS mutations: a systematic review and meta-analysis. Sci Rep. 2014;4:7521. -Assuming that additional mutations similar to the one shown in Figure 1 occur throughout the mutant K-ras gene, what would be the best explanation for the difference in denaturation between wild-type (WT) and mutant K-ras DNA?

Accepted Answer

11ecba2d_0e70_2d78_a3bf_ebe5591c340f_MD0008_00 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 (T_m) 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 T_m 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 T_m). (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 T_m. (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 T_m 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 (T_m) and stability of double-stranded DNA.

Question 16

Passage The K-ras gene codes for a small guanine nucleotide-binding protein, which acts as a molecular activator in numerous signaling pathways that promote DNA replication along with cell growth and survival. K-ras mutations at codons 12 and 13 function as biological markers for diagnosis of tumors. The DNA of SW-480 cells, a line originating from a colon adenocarcinoma, is the standard positive control used to identify codon 12 mutations in K-ras testing. Figure 1 Structural comparison of K-ras codon 12 in wild-type cells and SW-480 cellsPCR sequence analysis has been the preferred method for studying genetic mutations such as those in K-ras. However, challenges in analyzing PCR amplicons have prompted the development of DNA melting curve analysis (MCA). In the MCA technique, DNA is extracted from clinical samples and amplified by PCR. The amplified DNA is then melted (30-70°C) and analyzed by assessing the fluorescence (at 640 nm) of LC Green Plus, an intercalating dye that fluoresces when bound to double-stranded DNA (Figure 2). Figure 2 Melting curves of wild-type (WT) and SW-480 K-ras DNA.To study K-ras activity in colorectal cancer cells, short tandem repeats (STRs) were introduced into the K-ras gene. STRs are short stretches of DNA with repeating nucleotides, and four different STR sequences were inserted into the K-ras gene and analyzed via MCA under different experimental conditions (Table 1).Table 1 Sequences Studied in K-ras DNA MCA Analysis. Liu YP, Wu HY, Yang X, et al. Diagnostic accuracy of high resolution melting analysis for detection of KRAS mutations: a systematic review and meta-analysis. Sci Rep. 2014;4:7521. -What are the physical states of wild-type (WT) and SW-480 K-ras DNA at 55°C?

Accepted Answer

11ecba2d_0e6f_1c07_a3bf_c5aecf477c3f_MD0008_00 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 (T_m), 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 T_m 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 (T_m) is achieved when 50% of double-stranded DNA has become single-stranded; T_m is usually depicted as the midpoint of the melting curve.

Question 17

Passage Nerve cells must maintain the composition of their myelin sheaths to function properly. Galactocerebroside (GC), shown in Figure 1, is a hydrolyzable lipid and an important component of the myelin sheath. Figure 1 Structure of the myelin sheath component galactocerebrosideHomeostasis of GC is maintained by its dynamic synthesis and degradation, as shown in Figure 2. GC is synthesized by the enzyme ceramide galactosyltransferase (CGT) using UDP-galactose (UDP-Gal) and ceramide (Cer) as substrates. It is degraded by the enzyme galactocerebrosidase (GalC), which hydrolyzes the glycosidic bond to release galactose (Gal) and ceramide. A deficiency in GalC activity leads to a buildup of GC, and prior research has connected this buildup with neurodegeneration. Figure 2 Schematic of GCT and GalC activity in healthy and GalC deficient cellsOver 100 inactivating GalC mutations have been identified. Five of the most common mutations are listed in Table 1. The changes to both DNA and amino acid sequence are shown.Table 1 Five Common Galactocerebrosidase Mutations -After galactose is released from GC and linearized, it contains all of the following functional groups EXCEPT:

Accepted Answer

11ecba2d_0e64_468a_a3bf_1d07813e4b36_MD0008_00 Monosaccharides are chains of three or more carbon atoms with the general formula C_nH_2nO_n. 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.

Question 18

Passage The K-ras gene codes for a small guanine nucleotide-binding protein, which acts as a molecular activator in numerous signaling pathways that promote DNA replication along with cell growth and survival. K-ras mutations at codons 12 and 13 function as biological markers for diagnosis of tumors. The DNA of SW-480 cells, a line originating from a colon adenocarcinoma, is the standard positive control used to identify codon 12 mutations in K-ras testing. Figure 1 Structural comparison of K-ras codon 12 in wild-type cells and SW-480 cellsPCR sequence analysis has been the preferred method for studying genetic mutations such as those in K-ras. However, challenges in analyzing PCR amplicons have prompted the development of DNA melting curve analysis (MCA). In the MCA technique, DNA is extracted from clinical samples and amplified by PCR. The amplified DNA is then melted (30-70°C) and analyzed by assessing the fluorescence (at 640 nm) of LC Green Plus, an intercalating dye that fluoresces when bound to double-stranded DNA (Figure 2). Figure 2 Melting curves of wild-type (WT) and SW-480 K-ras DNA.To study K-ras activity in colorectal cancer cells, short tandem repeats (STRs) were introduced into the K-ras gene. STRs are short stretches of DNA with repeating nucleotides, and four different STR sequences were inserted into the K-ras gene and analyzed via MCA under different experimental conditions (Table 1).Table 1 Sequences Studied in K-ras DNA MCA Analysis. Liu YP, Wu HY, Yang X, et al. Diagnostic accuracy of high resolution melting analysis for detection of KRAS mutations: a systematic review and meta-analysis. Sci Rep. 2014;4:7521. -Which of the following is most likely to be identified in SW-480 cells?

Accepted Answer

11ecba2d_0e6d_bc76_a3bf_e5fd39e8d2c2_MD0008_00 Cancer cells are able to grow and divide excessively due to loss of cell cycle control. Various types of cancers result from oncogene activation, tumor suppressor gene inactivation, or both. Proto-oncogenes are wild-type (WT) genes that are normally involved in cell cycle progression (eg, cell division stimulation, apoptosis prevention). However, when proto-oncogenes are mutated into oncogenes via activating mutations in one or two alleles, cells become cancerous. As a result, oncogenes lead to either an overexpressed protein, a hyperactive protein, or both, promoting uncontrolled cell proliferation. Tumor suppressor genes (or antioncogenes) regulate DNA repair by repressing or pausing the cell cycle to ensure that only normal cells proceed to the division (mitosis) stage. These genes also prevent accumulation of mutations in cancer cells by either repairing the mutations or inducing programmed cell death (apoptosis) if repair fails. In the setting of cancer, tumor suppressor genes become inactivated by loss of function mutations in both alleles and lose the ability to prevent abnormal growth and division of damaged cells. According to the passage, the K-ras gene normally encodes a protein responsible for activating the signaling pathways that promote DNA replication (cell cycle S-phase) along with cell growth and survival. Mutated K-ras in SW-480 cells is most likely constantly inducing signaling pathways responsible for the characteristic uncontrolled growth, division, and survival of cancer cells. Therefore, K-ras is most likely a proto-oncogene with an activating mutation in at least one allele that turned it into an oncogene. (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.

Question 19

During glycolysis, phosphofructokinase converts fructose 6-phosphate (F6P) into fructose 1,6-bisphosphate (F1,6BP), as shown below.  Aldolase then converts F1,6BP into glyceraldehyde 3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP).  Respectively, phosphofructokinase and aldolase are examples of:

Accepted Answer

11ecba2d_0e74_251d_a3bf_3561c27e50ef_MD0008_00 Enzymes are classified into one of six categories based on their mechanism of catalysis. Transferases move a chemical group from one molecule to another. Kinases are a type of transferase that either move a phosphate group from ATP to a substrate to form ADP and phosphorylated substrate, or from a substrate to ADP to form ATP. Phosphofructokinase transfers a phosphate from ATP to fructose 6-phosphate (F6P), forming ADP and fructose 1,6-bisphosphate (F1,6BP). Therefore, phosphofructokinase is a transferase. Lyases either form two molecules by breaking a bond within one molecule or they combine two molecules into one. To form bonds between reactant molecules, lyases add groups to a double bond. In the reverse reaction, they break a single bond in a reactant and generate a new double bond in the products. Aldolase splits F1,6BP into glyceraldehyde 3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP) by forming a new C=O double bond. Therefore, aldolase is a lyase. (Choices A and C) Isomerases change the positions of functional groups within a single molecule, and hydrolases break bonds by adding water to them. Phosphofructokinase and aldolase transfer functional groups between molecules and break bonds by creating new double bonds, respectively. Ligases use the energy from ATP hydrolysis to combine two molecules into one. (Choice D) Oxidoreductases participate in redox reactions by transferring electrons from one molecule to another, and usually require NAD(P)H or FADH₂ as a substrate or a product. Neither phosphofructokinase nor aldolase catalyze redox reactions. Educational objective: Enzymes are classified based on their mechanism of catalysis. Oxidoreductases move electrons; transferases move functional groups between molecules; hydrolases break bonds by adding water, and lyases break bonds by forming a double bond elsewhere; isomerases move functional groups within a molecule; and ligases join molecules together using ATP as an energy source.

Question 20

Passage Nerve cells must maintain the composition of their myelin sheaths to function properly. Galactocerebroside (GC), shown in Figure 1, is a hydrolyzable lipid and an important component of the myelin sheath. Figure 1 Structure of the myelin sheath component galactocerebrosideHomeostasis of GC is maintained by its dynamic synthesis and degradation, as shown in Figure 2. GC is synthesized by the enzyme ceramide galactosyltransferase (CGT) using UDP-galactose (UDP-Gal) and ceramide (Cer) as substrates. It is degraded by the enzyme galactocerebrosidase (GalC), which hydrolyzes the glycosidic bond to release galactose (Gal) and ceramide. A deficiency in GalC activity leads to a buildup of GC, and prior research has connected this buildup with neurodegeneration. Figure 2 Schematic of GCT and GalC activity in healthy and GalC deficient cellsOver 100 inactivating GalC mutations have been identified. Five of the most common mutations are listed in Table 1. The changes to both DNA and amino acid sequence are shown.Table 1 Five Common Galactocerebrosidase Mutations -Based on the information in the passage, which bond is cleaved by GalC?

Accepted Answer

11ecba2d_0e65_30eb_a3bf_8b7af12f320f_MD0008_00 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.

Exam 2: Biochemistry

A solution of which amino acid will NOT rotate plane polarized light?

Proline differs from the other standard amino acids in that only the former:

If a stretch of double-stranded DNA has a total of 560 nucleotides and 159 adenine nucleotides, what is the nucleotide composition of this DNA molecule? It has:

Which factor decreases the stability of a DNA double helix during denaturation?

Which of the following molecules could decrease the activity of one or more Krebs cycle enzymes?

The compound 5,10-methylene tetrahydrofolate (5,10mTHF) is derived from folic acid (vitamin B9) and is required in the active site of the enzyme thymidylate synthase for thymine synthesis. Therefore, 5,10mTHF is most likely:

The pKa for the side chain of cysteine is 8.4. What is the ratio of protonated side chain (-SH) to deprotonated side chain (-S−) in a solution of cysteine at physiological pH of 7.4?

A fatty acid undergoes complete β-oxidation to yield nine molecules of acetyl-CoA. No isomerization reactions occur. The structure of the original fatty acid was:

An infant who experiences seizures and has lactic acid buildup is shown to have deficient pyruvate dehydrogenase complex (PDHC) activity. Genetic analysis shows that both the maternal and paternal alleles for all PDHC subunits are normal. These results could indicate:

Under physiological conditions, peptide bond formation and degradation both require enzymes, but only formation requires coupling to GTP hydrolysis. Based on this information, peptide bonds under physiological conditions are:

Which series depicts the order in which the precursors of steroid hormones are synthesized?

Based on the titration curve of an unknown amino acid, shown below, can the side chain of the amino acid readily exchange protons with water?

The compound 5,10-methylene tetrahydrofolate (5,10mTHF) is derived from folic acid (vitamin B₉) and is required in the active site of the enzyme thymidylate synthase for thymine synthesis. Therefore, 5,10mTHF is most likely:

The pK_a for the side chain of cysteine is 8.4. What is the ratio of protonated side chain (-SH) to deprotonated side chain (-S⁻) in a solution of cysteine at physiological pH of 7.4?