Biochemistry

question 22

** have to know the structures - sticky note in notebook** Enantiomers (D vs L) = same chem formula but different connectivity so different at every stereo center. Epimers = different at only 1 stereo center Anomers = special case of epimers that differ from each other at the anomeric carbon and they can only exist in cyclized sugars (furanose or pyranose) bc in the linear form the anomeric carbon is a carbonyl and cant be a stereo center A is alpha D-mannose C is beta D fructose D is beta-L glucose (enantiomer of beta D- glucose) Concept: 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.

Question 2

15N is being added to a media containing 14N. Generation 1 will be 15-14 so 100% of double helices formed will have 1 15N and 14N. Gen 2 will have 50% 15-14 and 50% 14-14. In Gen 3, you will get 25% double helices containing both. Ratios: Gen 1 = 100% Gen 2 = 50% Gen 3 = 25% Gen 4 = 12.3% Concept: 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.

Aldoses vs ketoses:

Alodoses - sugar has an aldehyde group (R-CHO), so carbonyl group (C=O) at end of the C chain Ketoses - sugar has a ketone group (R-C=O-R'), so carbonyl group is internal to the chain **looks like only fructose is a ketose and all the other sugars are aldoses**

question 26

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. 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. Concept: 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.

Question 3

A ribose is a 5 sugar molecule so it had to be btw A and B. The structure of pyranose and furanose is shown. 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. In the pyranose form, the 5′ hydroxyl group of ribose becomes part of the ring structure (look at the image) and cannot participate in phosphoester bonds, so ribose cannot be in the pyranose form in NTPs. Pyranose = 6 atoms Furanose = 5 atoms "I want more PIE (py) so pyranose has more atoms" Concept: 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.

question 18

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. Concept: 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.

question 39

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. concept: 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.

question 28

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. Concept: 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.

question 9

According to the passage, avian receptors contain α-2,3-glycosidic bonds whereas human receptors have α-2,6-glycosidic bonds. Therefore, neuraminidase action in humans is expected to increase the cleavage of SA and reduce the number of α-2,6-glycosidic bonds. Concept: Glycosidic bonds are created by the linkage of a nucleophilic functional group, such as a hydroxyl or amino group, to the anomeric carbon of a sugar molecule. The bond is classified based on the assigned numbers of the linked carbons and on whether the anomeric carbon is in the α- or the β-conformation.

question 3

According to the passage, they said that phosphorylated leptin decreases FA synthesis and increases beta oxidation. The question is asking what will happen when a phosphatase is added. The phosphatase would remove the phosphate, inactivating leptin and so it would increase FA synthesis and downregulate beta oxidation. Choice C is taking about the FAS complex and the acyl carrier protein in the FAS which is used for FA synthesis concept: In the cytosol, ACC catalyzes the first (rate-limiting) step of fatty acid synthesis by carboxylating acetyl-CoA to malonyl-CoA. Fatty acid synthase then carries out the main reactions of fatty acid synthesis. This enzyme contains an acyl carrier protein, which transfers acyl intermediates.

Question 15

According to the question, methionine enters the citric acid cycle by conversion to succinyl-CoA. At this point in the cycle, two oxidative steps remain. The first step is the conversion of succinate to fumarate, which produces one FADH2 molecule. The second step is conversion of malate to oxaloacetate, which produces one NADH molecule. All other oxidative steps in the cycle require precursors to succinyl-CoA, which do not form in the metabolism of methionine. Based on this, entry of methionine into the citric acid cycle will yield one NADH molecule. Concept: Different molecules can enter the citric acid cycle at different points. The cycle oxidizes these molecules to produce the reduced electron carriers NADH and FADH2. The number of electron carriers produced depends on the number of oxidative steps remaining in the cycle at the point of entry.

question 5

According to the table, addition of citrate increases the activity of ACC more. But the reason why its an allosteric effector and not a cofactor (metal ions) or coenzymes (small organic compounds) is bc without citrate present, ACC is still active but just at a lower level. Cofactors and coenzymes bind to the enzymes active site and are REQUIRED for enzymatic activity whereas allosteric effectors just regular enzymatic activity by binding at sites other than the active site (allosteric site) and causing conformational changes in the enzyme. Concept: Allosteric effectors regulate enzymatic activity by binding at sites other than the active site and causing conformational changes in the enzyme. As a result, enzymatic activity is either inhibited or activated (ie, increased).

question 21

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. Concept: 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.

question 33

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 donorswhereas cytosine contributes 2 acceptors and 1 donor. Concept: Hydrogen bond acceptors are electronegative atoms (nitrogen or oxygen) that have at least one lone pair of electrons, and hydrogen bond donors are hydrogen atoms bound to electronegative atoms. Watson-Crick base pairing occurs when particular hydrogen bond donors and acceptors in one nucleotide bond to acceptors and donors in the complementary nucleotide. Adenine and thymine each have 1 Watson-Crick donor and 1 acceptor, whereas guanine has 2 donors and 1 acceptor atom, and cytosine has 1 donor and 2 acceptors.

Question 14

Adipose tissue stores fat and so fat soluble vitamins would accumulate in adipose tissue. 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 Concept: 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.

question 23

Ans is B bc it is a steroid molecule and steroid molecules can cross the cell membrane and directly interact with their effector proteins inside the cell. Backbone of steroids consists of three 6-membered rings and one 5-membered ring fused together. Hydrophilic signal molecules such as peptide hormones interact poorly with the hydrophobic tails of phospholipid bilayers, so they cannot easily cross cell membranes. Instead, these hormones (first messengers) bind to cell surface receptors such as G-protein-coupled receptors. concept: 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.

question 44

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. Concept: 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.

question 42

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. concept: 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.

Question 11

DNA double helix stability depends on H bonds GC = 3 bonds AT = 2 bonds Concept: 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 50

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.

Question 54

During fatty acid synthesis, NADPH is oxidized to NADP+ to reduce the carbonyl groups and carbon-carbon double bonds on each acetyl-CoA molecule added to the fatty acid chain. If fatty acid synthesis is inhibited in ME patients, the conversion of NADPH to NADP+ will occur less frequently, and cytosolic NADPH will build up. Anabolic processes generally require energy, reducing power, and sufficient precursor molecules: in fatty acid synthesis these requirements are satisfied by ATP, NADPH, and acetyl-CoA, respectively. Concept: Acetyl-CoA, NADPH, and ATP are the reactants needed to generate fatty acid chains during lipid synthesis. The pentose phosphate pathway generates the NADPH needed to reduce the carbonyl groups from each molecule of acetyl-CoA that is added to a fatty acid chain.

Question 11

During glycolysis, NAD+ is reduced to NADH as glyceraldehyde 3-phosphate is oxidized to 1,3-bisphosphoglycerate; therefore, NAD+ is required for glycolysis to continue. Under aerobic conditions, NAD+ can be regenerated by oxidation of NADH in the electron transport chain (ETC). Under anaerobic conditions, the ETC cannot regenerate enough NAD+, and another source is required. Under these conditions, lactate synthesis takes place and oxidizes NADH to NAD+ as pyruvate (the final product of glycolysis) is reduced to lactate. concept: Lactate synthesis is coupled to the synthesis of NAD+ from NADH. Under anaerobic or oxygen-poor conditions, this process provides the NAD+ necessary to continue glycolysis.

Question 7

ETC generates protons which drive the phosphorylation of ADP into ATP Educational objective: In oxidative phosphorylation, a series of energetically favored redox reactions produces protons. The protons are pumped from the mitochondrial matrix to the intermembrane space, creating an electrochemical gradient that drives ATP synthesis.

question 10

FA acids modified/ activated to enter the matrix. Steps: 1. In the cytosol, the enzyme acyl-CoA synthetase catalyzes the reaction of fatty acids with coenzyme A to form acyl-CoA molecules. This reaction is thermodynamically unfavorable and requires ATP hydrolysis to proceed 2. Acyl-CoA molecules then migrate to the intermembrane space, where they can react with carnitine to form acylcarnitine. The transport protein acylcarnitine translocase, located on the inner mitochondrial membrane, recognizes acylcarnitine and carries it into the mitochondrial matrix 3. Acylcarnitine is then converted back to acyl-CoA and carnitine. Acyl-CoA is subsequently digested by β-oxidation, and carnitine is transported out of the matrix where it can pick up and carry a new fatty acid into the mitochondrial matrix OAA is not involved in transport of FA from the cytosol into the mito concept: Entry into the mitochondrial matrix is tightly regulated by transport proteins in the inner mitochondrial membrane. Fatty acids must be activated with coenzyme A followed by carnitine to enter the mitochondrial matrix. Activation requires ATP hydrolysis.

question 35

FA synthesis occurs in the cytosol and oxidation occurs in the mitochondria concept: Opposing catabolic and anabolic pathways are regulated hormonally, allosterically, and by compartmentalization. Fatty acid oxidation (degradation) and fatty acid synthesis are compartmentalized, with oxidation occurring in the mitochondria and synthesis occurring in the cytosol.

question 45

Fatty acid degradation to acetyl-CoA yields NADH and FADH2, both of which can enter the electron transport chain to produce ATP. In addition, the resulting acetyl-CoA can enter the citric acid cycle to produce more NADH and FADH2, as well as GTP (an ATP equivalent). Therefore, of the choices given, fatty acid oxidation most likely provides the energy needed for sustained gluconeogenesis. Glycogenolysis produces glucose, so when glycogen is present, gluconeogenesis is not required. Accordingly, gluconeogenesis generally occurs after glycogen stores have been depleted, so glycogenolysis would not be able to provide the needed energy. Concept: Gluconeogenesis is an anabolic process that requires energy input from ATP equivalents. The necessary energy is provided by catabolic processes, most commonly by fatty acid oxidation.

Question 16

Fatty acid synthesis is the process of linking acetyl-CoA units together and reducing carbonyls to the alkyl form. To synthesize a fatty acid containing 2n + 2 carbons, n ATP molecules and 2n NADPH molecules must be consumed, forming ADP and NADP+. The question asks how many ADP and NADP+ molecules will form during synthesis of a 16-carbon fatty acid chain. The scheme also shows that the resulting fatty acid contains 2n + 2 carbons. If the total number of carbons in the fatty acid chain (2n + 2) is 16, then 2n = 14 and n = 7. Therefore, 7 ADP molecules and 14 NADP+ molecules form in this process. concept: Fatty acid synthesis is the process of linking acetyl-CoA units together and reducing carbonyls to the alkyl form. To synthesize a fatty acid containing 2n + 2 carbons, n ATP molecules and 2n NADPH molecules must be consumed, forming ADP and NADP+.

question 12

If theres 159 adenine then there would be 159 thymine and the only choice with that option is C so you don't have to do it out Concept: 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 48

Figure 1 shows that L-carnitine and 5-AVAB have nearly identical structures, differing only at the hydroxyl (OH) group on L-carnitine and the extra methylene (CH2) group on 5-AVAB. OCTN2 transports L-carnitine into cells, and 5-AVAB is likely to compete for this interaction due to its similarity to L-carnitine. Moreover, Figure 2 show that 5-AVAB decreases the rate of oxygen consumption (likely due to reduced β-oxidation of fatty acids in the mitochondria), and the information in the passage indicates that this is likely due to decreased L-carnitine transport by OCTN2. Figure 2 also shows that increasing the concentration of L-carnitine from 50 μM to 2 mM overcomes the effect of 5-AVAB, indicating that the two molecules compete for interaction with OCTN2. Therefore, 5-AVAB most likely alters fatty acid metabolism by competition with L-carnitine for OCTN2 interactions. concept: Transport proteins facilitate movement of molecules across membranes. They may transport more than one type of molecule if the molecules have similar properties but can usually interact with only one molecule at a time. Similar molecules may compete for interactions with a transport protein.

question 49

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). concept: 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.

question 43

Figure 1 shows that in the presence of F2,6BP, the glycolysis enzyme PFK-1 has higher activity than in the absence of F2,6BP. Conversely, the gluconeogenesis enzyme F1,6BPase has lower activity in the presence of F2,6BP than in its absence. Therefore, PFK-1 is activated by F2,6BP and F1,6BPase is inhibited. But also, I knew that F2,6BP is a positive regulator for PFK1 Other allosteric regulators for PFK1: AMP (ADP) = positive allosteric ATP = negative allosteric Citrate = negative allosteric F2,6-BP = positive allosteric concept: Metabolic pathways are commonly regulated by allosteric effectors. Allosteric effectors bind enzymes at sites other than the active site and induce conformational changes that alter enzymatic activity. A molecule that allosterically activates one pathway often allosterically inhibits the opposing pathway.

Question 5

Figure 1 shows that the influenza A receptor is a glycoprotein because the carbohydrate chain is linked to a protein through the amino acid asparagine. The terminal group of the chain consists of SA bound to two, linked sugar molecules (a disaccharide). Both sugars in the disaccharide are bound to another molecule at their anomeric carbons, so neither contains a hemiacetal or hemiketal. Therefore, the disaccharide is a nonreducing sugar, and the correct description of the receptor should state that it is a glycoprotein with SA bound to a nonreducing disaccharide. A sugar can be classified as reducing when it has a free anomeric carbon for another monosaccharide to bind it. If it doesn't have it than its nonreducing Concept: 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.

Question 2

Figure one x axis shows the effect of of AMPK at .5 hr, 1hr 6hr (all i.v) and 1hr (i.h.p). The increased amount of time taken for PH to lower AMPK activity (ie, inactive at 0.5 hr but active at i.v. 6 hr and 1 hr i.h.p) suggests that PH acts in a time-dependent manner. From the passage, they explained the pathway which is why pH is an antagonist Concept: An agonist enhances or duplicates the effect of a ligand binding to a receptor. Antagonists alter receptor-ligand binding affinity or bind directly to a receptor to decrease the cell's response to the ligand. The effect of agonists or antagonists can be time-dependent (ie, may not be immediate).

question 55

First, fatty acids in the cytoplasm are activated when the enzyme acyl synthetase attaches them to the carrier molecule CoA using ATP. Subsequently, in the rate-limiting step of the reaction, carnitine palmitoyltransferase I (CPTI) converts acyl-CoA molecules into acylcarnitine, which enters the intermembrane space. Acylcarnitine is then moved by acylcarnitine translocase across the inner membrane and into the matrix. Lastly, to begin the oxidation reactions, carnitine palmitoyltransferase II (CPTII) on the inner membrane reconverts fatty acylcarnitine into fatty acyl-CoA, which can be broken down into acetyl-CoA. Therefore, researchers interested in confirming that fatty acid oxidation is not impaired would need to ensure that translocation of acylcarnitine into the mitochondrial matrix is functional. Concept: During β-oxidation, long-chain fatty acids are activated with coenzyme A and shuttled by enzymes from the cytoplasm to the mitochondrial matrix. The rate-limiting step of fatty acid oxidation is the conversion of fatty acyl-CoA molecules into fatty acylcarnitine by carnitine acyltransferase

question 47

From the pathway they mentioned in the paragraph, F2,6BPase acts as glucagon to convert F2,6BP into F6P. So, if glucagon levels are high you will get lower levels of F2,6BP (look at image) concept: The hormones insulin and glucagon regulate blood glucose levels. When blood glucose is high, the pancreas releases insulin, which upregulates glucose uptake, glycolysis, and glycogen synthesis. When blood glucose is low, the pancreas releases glucagon, which stimulates gluconeogenesis and glycogenolysis to release glucose from the liver into the bloodstream.

Common monosaccharides to disaccharides

Glucose + Fructose = Sucrose (a-1,2) Galactose + Glucose = Lactose (b-1,4) Glucose + Glucose = Maltose (a-1,4) The anomeric carbon is the carbon that is bonded to 2 Oxygen atoms, and is the highest priority group for naming. We consider it an ALPHA glycosidic bond if it's anomeric carbon is bonded axial downwards. Beta is the opposite, where its bonded upwards and equatorial. ("Look up its a bird" - beta is up and alpha is down )

question 9

Glycogenolysis is the breakdown of glycogen into glucose. It is the first source of glucose during fasting and requires G6Pase in its final step (Number I). Gluconeogenesis is the synthesis of glucose from pyruvate and other precursors. It is essentially glycolysis in reverse with a few differences, including the conversion of pyruvate to oxaloacetate and the use of phosphatases such as G6Pase instead of kinases to remove phosphate groups (Number II). (Number III) The pentose phosphate pathway is required for nucleotide and NADPH synthesis. Although the pathway includes G6P as a substrate, it does not require G6Pase activity, so it will not be directly impaired by von Gierke disease. The NADPH thats produced by this pathway feeds into fatty acid synthesis Concept: Several biological processes, including gluconeogenesis, glycogenolysis, and the Cori cycle, help buffer blood glucose levels. All of these processes require glucose 6-phosphatase (G6Pase) to catalyze the final step to release free glucose. The pentose phosphate pathway uses glucose 6-phosphate as a substrate, but does not produce free glucose or require dephosphorylation and is unaffected by changes in G6Pase activity.

question 44

Glycolysis is a catabolic process that degrades glucose to pyruvate. For every glucose molecule that completes glycolysis, a net total of two ATP molecules are produced (2 consumed and 4 generated). In contrast, gluconeogenesis is an anabolic process. According to the passage, gluconeogenesis requires the net input of 6 ATP equivalents (2 GTP and 4 ATP) to produce one glucose molecule. Glycolysis in the muscles is connected to gluconeogenesis in the liver by the Cori cycle.In this scenario, one glucose molecule is converted to two pyruvate molecules in the muscle, which are then sent to the liver (as lactate), where they are converted back to glucose. Because gluconeogenesis consumes 6 net ATP equivalents and glycolysis produces 2 net ATP equivalents, a net total of 4 ATP equivalents are consumed in one round of the cycle. concept: Catabolic processes lead to energy production, typically in the form of ATP equivalents. In contrast, anabolic processes consume energy. Glycolysis (a catabolic process) and gluconeogenesis (an anabolic process) are connected by the Cori cycle.

question 58

Had to know the process of protein catabolism for this question. Protein catabolism in the liver refers to the breakdown of polypeptide chains and proteins into individual amino acids, which can be further metabolized to enter the citric acid cycle and produce ATP. This requires a transamination reaction that removes the amino group (NH3+) from the carbon backbone of the amino acid. The transamination reaction produces an α-keto acid from the amino acid by transferring the NH3 group to α-ketoglutarate, forming glutamate. The glutamate produced in this process is then deaminated, releasing ammonia (NH3), which enters the urea cycle. The urea formed in this cycle is released in urine. According to the passage, proteins in male patients are catabolized at a faster rate than females. As a result, male patients would be expected to produce a higher concentration of glutamate in their liver cells compared with female patients. Concept: Protein catabolism refers to the breakdown of polypeptide chains and proteins into individual amino acids to produce ATP, glucose, or new proteins. Transamination reactions generate α-keto acids from amino acids by transferring the NH3 group to α-ketoglutarate, which is converted to glutamate.

Question 59

Had to recognize that glycine and cysteine are glucogenic AAs not ketogenic meaning that they will get converted to pyruvate and pyruvate can become two things: 1) glucose - through gluconeogenesis 2) lactate - through fermentation Ans is not glucose bc conversion of pyruvate to glucose occur sin gluconeogenesis which requires ATP. Aerobic respiration is required to produce sufficient ATP for gluconeogenesis. Qstem states "extended periods of physical exertion" meaning that we are in anaerobic respiration. Ketogenic amino acids are converted directly to acetyl-CoA, which can enter the citric acid cycle or be used to form ketone bodies. The passage uses this classification to place amino acids into the three categories in Table 1. Amino acids in Category I are converted exclusively to pyruvate, and those in Category II become acetyl-CoA. Category III consists of amino acids used to make various citric acid cycle intermediates. Glucogenic amino acids are converted to pyruvate or citric acid cycle intermediates, which can be converted to glucose. concept: Glucogenic amino acids are converted to pyruvate, which can be used to make glucose or citric acid cycle intermediates. Ketogenic amino acids are converted directly to acetyl-CoA, which can enter the citric acid cycle or be used to form ketone bodies. In the absence of oxygen, pyruvate is reduced to lactate to regenerate NAD+ during fermentation.

Question 10

I found that answer because in the passage it says the envelop contains phosphatidylethanolamine and neg charged phospholipids. So, I guessed phosphatidylserine. Also, it is a negative charged phospholipid. Phosphatidylserine has a negative charge Concept: 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.

Question 15

I looked at the diagram and figured out the right answer. But in the passage they say D-mannose connects to serine through an α-linkage indicating that the linkage is through the anomeric carbon (carbon 1) in the α-conformation. D-mannose connects to the next sugar in the chain in a β-1,4 linkage, which indicates a bond between the anomeric carbon of a sugar in the β-conformation and carbon 4 of another sugar. Because the anomeric carbon of D-mannose is already linked to serine, carbon 4 of D-mannose must be linked to the anomeric carbon of the next sugar in the chain. Finally, POMK attaches a phosphate group to D-mannose at carbon 6 "Look up its a bird" - beta is up and alpha is down Concept: Carbohydrate carbons are numbered starting at the end of the linear chain closest to the anomeric carbon. Most sugars can exist as linear or cyclical molecules and upon cyclization can adopt either an α- or a β-conformation at the anomeric carbon.

question 40

I picked it bc there was no phospholipid in the structure but all the other choices had it. A lipid with a phosphate group modification is called a phospholipid, and a lipid with a carbohydrate attached is called a glycolipid. concept: 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 17

If the partial pressure of oxygen in the environment were increased, yeast cells would exhibit increased function of the ETC and the need for fermentation to regenerate NAD+ would decrease. Because fermentation in yeast generates ethanol, increased partial pressure of oxygen (and decreased fermentation) would most likely result in reduced ethanol production. Concept: Glycolysis consumes NAD+ as a substrate. NAD+ can be regenerated by oxidizing NADH in the electron transport chain in the presence of oxygen, or by fermentation in the absence of oxygen. Fermentation converts pyruvate to lactate in higher eukaryotes, and to ethanol in bacteria and lower eukaryotes such as yeast.

question 4

In the passage it says that leptin increases beta oxidation. So, according to the question, leptin resistance would decrease beta oxidation meaning it would decrease acetyl coa production. Concept: Leptin increases fatty acid oxidation and decreases fatty acid synthesis. In contrast, leptin-resistant cells cannot increase fatty acid oxidation despite adequate leptin signaling. As a result, AMPK signaling cannot be activated by leptin and leads to increased and uninhibited ACC activity, promoting fatty acid synthesis and reducing β-oxidation.

question 56

In the passage they say that scientists predicted that the increased expression PDK1 will decrease PDH activity. According to figure 2. only PDK1 mRNA levels are significantly higher in ME patients. From the name you can tell that PDK1 has a kinase so it works but adding phosphate groups. So, the decrease in ATP levels in patients with ME is most likely bc of phosphorylation of PDH by PDK1. Concept: The pyruvate dehydrogenase complex catalyzes the decarboxylation reaction that converts pyruvate into acetyl-CoA. This reaction requires CoA, thiamine pyrophosphate, lipoic acid, and NAD+ and links glycolysis to the citric acid cycle. These pathways are essential for maximal ATP synthesis.

Question 6

In the passage they state that in the P186L mutation goes thru a nucleotide change of CCC to CUC. So, since the uracil is being added, ans choice D made the most sense. In the backbone of these strands, adjacent ribose sugars are joined by phosphodiester bonds at the 5′ and 3′ carbons. Each nucleotide contains a nitrogenous base bound to a sugar through a covalent bond known as a glycosidic bond. Concept: 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.

question 34

Just need to know the pentose phosphate pathway concept: The pentose phosphate pathway generates ribose 5-phosphate and NADPH. The NADPH is generated by reduction of NADP+, which is catalyzed by the oxidoreductases glucose 6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase.

question 45

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). concept: 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).

Explain what type of glycosidic bond the following forms: Maltose Lactose Sucrose Glycogen Starch Cellulose

Maltose - alpha 1,4 Lactose - beta 1,4 Sucrose - alpha 1,2 Glycogen - alpha 1,4 with alpha 1,6 branch Starch - alpha 1,4 Cellulose - beta 1,4

question 16

Mannose is an epimer of glucose because you simply switch the orientation at one chiral carbon Molecules with the same molecular formula but different arrangements of the atoms are called isomers. Stereoisomers are molecules with the same kinds of bonds at any given position but with the bonds oriented differently in space. Stereoisomers have one or more stereocenters, also called chiral centers, which are atoms bonded to four different chemical groups. Molecules that have multiple stereocenters but differ at only one of them are called epimers. D-mannose and D-glucose each have several stereocenters but differ from each other only at the carbon 2 center; therefore, they are called C-2 epimers. Because POMT can use activated D-mannose but not D-glucose as a substrate, it must be able to distinguish between these epimers. (Choice A) Unlike stereoisomers, constitutional isomers have the same molecular formula but different bond types in at least one position (ie, different connectivity between atoms). For example, D-glucose and D-fructose are constitutional isomers, having a carbon with two bonds to oxygen at carbon 1 and carbon 2, respectively. Educational objective: Stereoisomers are molecules that differ from each other in the spatial orientation of their bonds at one or more stereocenters (atoms that are bonded to four different chemical groups). Stereoisomers that differ at only one of several centers are epimers whereas those that differ at all centers are enantiomers. Constitutional isomers have the same molecular formula but distinct bond arrangements.

question 13

Mice in group A were not given tetracycline at any point and so have functional G6Pase. They therefore will be able to use any metabolites that can enter gluconeogenesis to help sustain their blood glucose levels. Amino acids (except for leucine and lysine) can be converted into glucose through gluconeogenesis. For example, alanine can be converted to pyruvate by deamination. Mice in group A will be able to use alanine to make glucose in the liver, thereby maintaining blood glucose levels. Mice in group B were given tetracycline at day 10 and will be unable to carry out gluconeogenesis after that point. Therefore, glucogenic substrates such as alanine will not help maintain their blood glucose levels. Concept: Blood glucose levels can be buffered by gluconeogenesis. Most amino acids are glucogenic, with the exception of leucine and lysine. Individuals who cannot carry out gluconeogenesis cannot buffer blood glucose with these metabolites.

question 36

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. concept: 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 34

Normally, DNA polymerases are equipped with both 5′-3′ and 3′-5′ exonuclease activity that allows them to remove and replace incorrect nucleotides at either end of a DNA strand. However, the passage states that the Klenow fragment (KF) enzyme described in the experiment does not have 5′-3′ exonuclease activity. KF can only proofread DNA in the 3′-5′ direction on the template strand, so only errors at the 3′ end of the growing strand can be repaired. Nucleotide repair enzymes are required to excise thymine dimers. Damaged bases are removed after replication by base excision repair enzymes with endonuclease activity. Concept: DNA polymerase I proofreads DNA and normally has exonuclease activity in the 5′-3′ as well as the 3′-5′ direction that allows it to remove primers and damaged or incorrect bases at the ends of the strand. It cannot fix mistakes in the middle of a strand; instead, base excision repair and nucleotide excision enzymes have endonuclease activity to remove damaged bases and mismatched nucleotides from the middle of a DNA strand, respectively.

question 50

OCTN2 transports L-carnitine into the cell which is required for the transport of FA from the cytosol and into the mito. So, silencing this would impair transport of FA into mitochondria which would impair their degradation so beta oxidation wouldn't happen and you wouldn't get acetyl-coa. Picture shows the structure for acetyl coa. concept: β-oxidation of fatty acids occurs in the mitochondria of eukaryotic cells and produces acetyl-CoA. L-carnitine is required for fatty acids to enter the mitochondria.

question 29

One of the things that contributes to membrane fluidity is double bonds (unsaturation level). So, for the answer choices, A and B are wrong bc you there is no double bond. Now comparing C and D, both have double bonds but you have now compare the fatty acid chain between them. When you have two unsaturated molecules, the shortest fatty acid chain will have fewest interactions and the greatest fluidity compared to the other one. Thats why answer is D bc the FA acid is lower in that. Concept: 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.

question 46

Passage stated that statins inhibited cholesterol precursor. Testosterone is a steroid and only one that would make sense Concept: 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.

question 38

Q is asking after galacatose is released from GC and becomes a straight chain which FG's will it have? Passage states that GalC enzyme hydrolyzes glycosidic bonds to release galactose and ceramide. So that bond breaks off galactose and within galactose the bond btw the O in ring and anomeric carbon breaks and causes an alcohol as side chain and the OH becomes a carbonyl bc its labeled as C1 cuz numbering starts with anomeric carbon. **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.** 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 1

Since the question is asking about gel electrophoresis, we need to look at the size of the nucleotides bc gel electrophoresis separates molecules by molecular weight with larger molecules migrating more slowly than smaller molecules. Guanine is heavier than adenine bc it has a carbonyl while adenine has the amine and O is heavier than N Concept: 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)

Question 1

Since the question is asking about the catabolic fate, the fatty acid chain would enter beta oxidation to be broken down into acetyl coa and the glycerol part would feed into glycolysis. The glycerol is phosphorylated to glycerol 3-P by glycerol kinase and it would feed into step 6 of glycolysis Concept: Triglycerides are storage lipid molecules that are broken down (catabolized) by lipases into glycerol and three fatty acids. Glycerol is converted into DHAP and incorporated into glycolysis or gluconeogenesis (anabolic pathway), depending on the cell's needs. The fatty acids diffuse into the mitochondria for β-oxidation, their major catabolic pathway.

question 14

Succinate-ubiquinone reductase, also known as Complex II or succinate dehydrogenase, is an oxidoreductase found in the inner membrane of the mitochondria in eukaryotes. In the electron transport chain, Complex II facilitates the transport of electrons from succinate in the citric acid cycle to Complex III. Electrons move through Complex II by three consecutive reactions. First, electrons from succinate are transferred to flavin adenine dinucleotide (FAD) to produce FADH2 as succinate is oxidized to fumarate. Then, FADH2 is reoxidized to FAD when electrons are shuttled to the iron-sulfur centers in Complex II. Finally, a small mobile electron carrier called ubiquinone, or coenzyme Q, accepts electrons from the iron-sulfur centers and transports them to Complex III. The question indicates that atpenins are effective antifungal agents that inhibit succinate-ubiquinone reductase (ie, Complex II). The transfer of electrons from Complex II to III through ubiquinone is the only reaction accurately described. Concept: Complex II (succinate-ubiquinone reductase or succinate dehydrogenase) oxidizes succinate to fumarate and transfers electrons to Complex III. Electrons are transferred from flavin adenine dinucleotide (FAD) to iron-sulfur centers in Complex II, then to ubiquinone (coenzyme Q), which carries them to Complex III through the inner membrane.

question 36

Table 1 showed which base pair was exchanged for what and it was a purine (A) for pyrimidine (T) concept: 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).

Question 13

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. Concept: 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 43

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 sphingosine with 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 concept: 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.

question 11

The cori cycle connects glycolysis and gluconeogenesis. Glycolysis produces lactate when less O2 available and if lactate builds up in muscles it gets sent to cori cycle to be converted back into glucose through gluconeogenesis concept: During anaerobic exercise, pyruvate is reduced to lactate to regenerate NAD+. Lactate that builds up in muscles is sent to the liver, where it is converted back to glucose and returned to muscles. This process is called the Cori cycle.

question 46

The data in Figure 1 confirm that PFK-1 is most active when F2,6BP is present, indicating that glycolysis is most active when F2,6BP is abundant. Therefore, conditions that promote glycolysis most likely also promote F2,6BP synthesis. Because the PFK-2 domain produces F2,6BP, it must be active during glycolysis. The passage states that the PFK-2 domain of the bifunctional enzyme is active (and the F2,6BPase domain is inactive) when it is dephosphorylated. Therefore, when glycolysis is active, the PFK-2/F2,6BPase enzyme is most likely dephosphorylated and produces F2,6BP concept: The rate of glycolysis is controlled by PFK-1, which is stimulated by the allosteric effector F2,6BP. When glycolysis is active, F2,6BP synthesis is also most likely active.

question 10

The disease in the passage inactivated gluconeogenesis. So, had to look for an enzyme thats involved in gluconeogenesis. Phosphoenolpyruvate carboxykinase (PEPCK) catalyzes the second step in gluconeogenesis, the conversion of oxaloacetate to phosphoenolpyruvate, and its inactivation can indeed lead to both lactate buildup and glucose depletion in the blood. concept: Inactivating one enzyme in a metabolic pathway often results in similar physiological consequences to inactivating another enzyme in the same pathway. Products downstream of the inactivated enzyme will not be produced, and upstream molecules will accumulate.

question 12

The disease mentioned in the passage said that it causes hypoglecemia which is low blood sugar. So, had to look for an answer that would give you sugar. Starch is a polysaccharide composed of several linked glucose molecules and its broken down into the glucose molexules in the digestive tract by amylase enzyme. This would help with the disease. In the passage they also say that the disease causes loss of sensitivity to hormones and thats why Epinephrine is the wrong choice ( I picked epi bc it has the same effects as glucagon) choice B: vitamin A is req for vision, immune system maintenance and growth concept: Genetic mutations that inhibit gluconeogenesis and glycogenolysis can prevent effective hormonal control of blood glucose levels. Dietary supplementation with a glucose source such as starch can help maintain glucose levels in these cases.

question 19

The easier way to get the answer is by looking at only where you see results which is in the p32 labeled ATP. Analyzing that autoradiogram, you can see that you see the results only when POMK is present and thats ans choice B. Reinforcing this you can compare to the ribitol 5 phosphate autoradiogram where there is not fukutin activity and that bc there is no POMK present. This tells us that fukutin needs POMK to be active Concept: 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.

question 52

The final step in the ETC is reduction of oxygen to form water, so the rate of oxygen consumption serves as a measure of ETC activity. The graph in Figure 2 shows that oxygen consumption is lowest in the presence of 100 μM 5-AVAB and 50 μM L-carnitine. Under these conditions ATP will be synthesized at a slower rate than under any other conditions tested, and ADP will be consumed at a slower rate. Therefore, the ADP/ATP ratio will be highest in cells at 100 μM 5-AVAB and 50 μM L-carnitine. concept: Aerobic organisms primarily generate ATP using energy provided by the electron transport chain (ETC). The final step in the chain is reduction of oxygen to water, and the rate of oxygen consumption can be used as a measure of ETC activity. Higher ETC activity corresponds to a higher ATP/ADP ratio or a lower ADP/ATP ratio.

Question 7

The graph is showing melting point difference between the mutations. H6-G225D has a lower melting point meaning its less stable and has a greater chance of denaturing. Concept: 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.

Question 6

The ligase class of enzymes catalyzes addition or synthesis reactions between two molecules by coupling them with the hydrolysis of energy-rich bonds such as those between ATP phosphate groups. Pyruvate carboxylase catalyzes the energy-requiring carboxylation of pyruvate to oxaloacetate, an intermediate for both amino acid synthesis and the citric acid cycle. Because pyruvate carboxylase "joined" pyruvate and CO2 to form oxaloacetate, the enzyme is classified as a ligase. Choice A: transferase Choice B: Lyases Choice C: Oxireductases Concept: The ligase class of enzymes catalyzes addition or synthesis reactions between two molecules by coupling them with the hydrolysis of energy-rich bonds such as those between ATP phosphate groups.

Question 4

They are asking for the most stable and when it comes to nucleotides, GC pair is more stable bc they have 3 hydrogen bonds as opposed to 2 H bonds in the AT pair. 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. (look at image for clarification) Concept: 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.

question 51

The passage states that palmitate (a fatty acid) was the only energy source given to cardiomyocytes, and therefore β-oxidation is the only pathway by which these cells can generate NADH and FADH2. Because β-oxidation is the only pathway that leads to the production of molecules that can enter the ETC in these cells, the rate of oxygen consumption is directly related to the rate of β-oxidation. Figure 2 shows that as the concentration of 5-AVAB increases, the rate of oxygen consumption decreases. Because palmitate is the only energy source, these data indicate that 5-AVAB reduces β-oxidation in a dose-dependent manner. concept: β-oxidation is the degradation of fatty acids to produce NADH and FADH2. The electrons from NADH and FADH2 are used to reduce oxygen to water. Accordingly, oxygen consumption can be used as a measure of electron transport chain activity.

question 27

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. II - triacylglycerides are responsible for energy storage III - sphingolipids only contain only one fatty acyl chains so it can only produce one fatty acid upon hydrolysis not 2 concept: 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.

Question 8

The question is asking about control so had to find the positive and negative control in the choices. I is right bc it serves as a negative control because it shows the expected response if the virus did not switch target species. II is right bc it serves as a positive control where you see the expected response if the mutant were to infect the expected cells aka the human cells here. III is wrong bc it doesn't infect the target species so it doesn't do us any good Negative control - where you expect nothing to happen so it acts like baseline for you to compare to Positive control - where you expect the effect to show up Concept: 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.

question 53

The question is saying that 5-AVAB might interfere with the ETC directly. To test this you can take two sets of cells where one wouldn't have 5-AVAB and one would. We can compare the ETC activity for these two sets of cells and if the set with the 5-AVAB is different from the set without it, you can conclude that 5-AVAB effects the activity of ETC directly. concept: The electron carriers needed for the electron transport chain (ETC), NADH and FADH2, can be provided by several pathways including glycolysis, β-oxidation, and the citric acid cycle. A molecule that affects the ETC directly will block the use of electron carriers regardless of which pathway generates them.

question 33

The question is saying that a mutation in ACADM gene causes MCAD deficiency (enzyme involved in beta oxidation). So MCAD deficiency will also lead to beta oxidation decrease meaning less acetyl coA production. So, a diet rich in glucose is an effective therapy because, like β-oxidation, digestion of glucose leads to the production of acetyl-CoA, reduced electron carriers, and ultimately ATP. concept: Inborn errors of metabolism are genetic disorders that result in impaired metabolic pathways. These disorders are typically treated either by supplementing the deficient enzyme with a functional version or by providing an alternate source of the desired metabolic products. Alternate sources must follow a separate, independent metabolic pathway from the defective one.

question 9

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. All fatty acids are broken down into multiple two-carbon units called acetyl-CoA. - FA w/ even number of carbons yield half as many acetyl-CoA units as the total number of carbons in the original molecule. - FA w/ 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. (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. Concept: 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 20

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. So, the western blot was performed to confirm Dg presence and it served as the positive control concept: 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

question 49

They say in the passage that L-carnitine moves INTO the cell through OCTN2 AGAINST its concentration gradient so that indicates active transport. The passage also says that the energy for this to happen comes from the transport of Na+ INTO the cell. This indicates symport. Concept: Many molecules enter and exit cells through transport proteins. Molecules may travel down their concentration gradient (passive transport, no energy required) or against their concentration gradient (active transport, energy required). Transport proteins facilitate uniport if they transport one molecule, symport if they transport two molecules in the same direction, and antiport if they transport two molecules in opposite directions.

question 42

To bypass the irreversible steps of glycolysis, gluconeogenesis uses different enzymes than those used in glycolysis to catalyze alternate reactions. According to the passage, PFK-1 transfers a phosphate from ATP to F6P during glycolysis to form ADP and F1,6BP. The reverse reaction would transfer the phosphate back to ADP, regenerating ATP and F6P. However, this reaction is too thermodynamically unfavorable to occur. Instead, gluconeogenesis uses F1,6BPase to hydrolyze the phosphate group from F1,6BP, forming F6P and an inorganic phosphate (Pi), not ATP. Therefore, F1,6BPase does not catalyze the reverse reaction of PFK-1. Also, in the passage they say "During gluconeogenesis, F1,6BPase removes the phosphate group by hydrolysis" concept: Metabolic pathways consist of both reversible and irreversible reactions. Opposing metabolic processes such as glycolysis and gluconeogenesis typically use the same enzymes for reversible reactions (going in opposite directions) but must use different enzymes to catalyze distinct reactions for the irreversible steps.

Sugars

Usually exist in D form. D form also means that the bottom carbon in a Fischer-Diagram is facing to the right. Atoms facing right in Fischer - drawn axially downwards in cyclic form Atoms facing left in Fischer - drawn to upwards and equatorially in cyclic form. For remembering structures, use the hand diagram Whichever way your fingers are pointing = where -OH bonds are on the fischer diagram D-Glucose: take right hand, face fingers left: middle finger faced outward. "fu*k glucose" D-Galactose: take left hand, make a spiderman fingers facing right. "In a parallel universe, spiderman protects the galaxy" D-Mannose: take right hand, face fingers left: two fingers faced outward. "shoot the man" D-Ribose: take right hand, face fingers left: make a thumbs up. "all's good for ribose" (note: must remember ribose and fructose are five carbon sugars) D-Deoxyribose: ribose but your index finger is MIA, literally "deoxy"d an -OH away D-Fructose: take mannose, but you lose one of your finger chambers. (Edit: usually a furanose but still a six carbon sugar)

question 25

You had to look at figure 1. 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. Concept: 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.

question 32

according to the passage, dFTP is an analog the lacks the Watson-crick hydrogen bonding and according to figure 1 there was no significant difference btw any of the nucleotides so that means dFTP is still binding to relatively the same amount as dNTP and thats why B is correct bc Watson crick H bonds are not required for base pair selectivity concept: 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.

question 35

dFTP dont have h bonds so the melting temp would be effected aka decrease Concept: 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.

question 37

first of all you have to realize that D galactose does have multiple stereocenters. You have to pick the highest number of stereocenter and do the R/S for that. If its R then it is D and if its S then it is L. Concept: 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

Question 17

galactose and ribose are monosaccharides so they def have a free anomeric carbon Even tho lactose is a dissaccharide, it has a free anomeric carbon. So does maltose - these two are reducing disaccharides Concept: Glycosidic bond formation requires a sugar with a free anomeric carbon, called a reducing sugar. All free monosaccharides are reducing sugars whereas disaccharides are reducing only if one anomeric carbon does not participate in a glycosidic bond. Sucrose is the most common nonreducing sugar.

Question 8

glucose 6-phosphatase (G6Pase) inactivation leads to decreased blood sugar and increased blood acidity. Group B mice were given tetracycline, which inactivated G6Pase 10 days into the experiment. Therefore, group B mice should maintain nearly constant glucose and pH levels prior to day 10, and should demonstrate decreased blood glucose and increased acidity after day 10. Because increased acidity means a decrease in pH, the correct graph should show a decrease in both glucose and pH levels, starting at day 10. Concept: Blood metabolite and pH levels are tightly controlled under normal conditions but can be altered by inactivation of metabolic enzymes such as glucose 6-phosphatase. An increase in acidity corresponds to a decrease in pH.

question 30

had to look in the passage and see that the base substitution was C to A so the answer choice had to be adenine (D) concept: 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

question 24

inflammatory responses occur near the site of injury. Arachidonic acid is then modified into prostaglandin which act as autocrine (cell signaling itself) and paracrine (cell signaling nearby cells). Since they act on themselves or on nearby cells, they cause localized inflammation. NSAIDs like aspirin and ibuprofen treat inflammation by inhibiting prostaglandin synthesis. Concept: 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.

question 48

looked at the graph and understood the melting temperature Concept: 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.

question 47

ras is an oncogene so thats why A is correct. I picked C but that wouldnt make sense bc its cancerous so the expression would be increased not decreased 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. Concept: 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 41

the ans is unsaturated cis as opposed to unsaturated trans bc trans is more stable than cis so cis would make the membrane more fluid (look at the image) Things that contribute to fluidity: 1. cholesterol level 2. saturation level -> unsaturated = more fluid (cis unsaturated > trans unsaturated) 3. # of C is FA chain -> more C = less fluid concept: 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.

question 57

the question is asking in which group would the have higher lipid synthesis. C is correct bc insulin would produce pyruvate which would produce acetyl Coa thru PDH and that acetyl coa would go to the cytosol to feed into FA synthesis. Since, in healthy individuals PDH is not effected, they would have a higher rate of lipid synthesis Choice B: chylomicrons are proteins and lipids that are digested by the enzyme lipoprotein lipase. This had nothing to do with the passage. Choice D: Insulin inhibits the enzyme hormone-sensitive lipase in adipocytes (fat cells) which breaks down triacylglycerides into fatty acids and glycerol. Therefore, few fatty acids would be able to contribute to lipid synthesis. Concept: Insulin stimulates lipid synthesis by activating the major enzymes involved in fatty acid production: Pyruvate dehydrogenase complex, acetyl-CoA carboxylase, and fatty acid synthase. PDH converts pyruvate produced from glucose into acetyl-CoA, which is transported to the cytoplasm for fatty acid synthesis.

question 8

the question states that the infant has a deficiency in PDH complex. Answer is B bc lipoic acid is a cofactor used in the complex. So, impaired lipoic acid production would result in decreased activity of the PDH complex. Function of PDH complex - 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. concept: Many enzymes rely on non-amino acid components called cofactors or coenzymes to function. The pyruvate dehydrogenase complex requires lipoic acid for activity.

question 31

the reason why C is right is bc in the passage they say that dFTP is a derivative of dNTPs. dNTPs is DNA used in our body so it doesn't have an OH in its 2' position and thats why dFTP would also lack that. Its not b bc they say in the passage that thymine bases is exchanged with difluototoluene in dFTP concept: 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.