MCAT Biochemistry Concept Checks and Missed Questions

Which of the diastereomers of glucose from the previous question are epimers of glucose? Enantiomers?

• Epimers: D-mannose (C-2), D-allose (C-3), D-galactose (C-4). • Enantiomers: None of the D-stereoisomers is an enantiomer for glucose; L-glucose is the enantiomer of D-glucose.

1. Describe the role of flippases and lipid rafts in biological membranes.

• Flippases: responsible for the movement of phospholipids between layers of the plasma membrane because it is otherwise energetically unfavorable. Lipid rafts: aggregates of specific lipids in the membrane that function as attachment points for other biomolecules and play roles in signaling

2. Given the following data, calculate the resting membrane potential of this cell: Ion: Na+ Permeability (Relative): 0.05 Intracellular Concentration: 14 mM Extracellular Concentration: 140 mM Ion: K+ Permeability (Relative): 1 Intracellular Concentration: 120 mM Extracellular Concentration: 4 mM Ion: Cl- Permeability (Relative): 0 Intracellular Concentration: 12 mM Extracellular Concentration: 120 mM

-64.0 mV or about -60 mV

3. What are the three major posttranscriptional modifications that turn hnRNA into mature mRNA?

1. 5'cap: addition of a 7-methylguanylate triphosphate cap to the 5' end of the transcript 2. 3' poly-A tail: addition of adenosine bases to the 3' end to protect against degradation 3. Splicing: removal of introns, joining of exons. Uses snRNA and snRNPs in the spliceosome to create a lariat, which is then degraded. Exons are ligated together.

3. What are the three major structural differences between DNA and RNA?

1. DNA is deoxyribose and has the 3' OH group replaced with -H. 2. DNA is double-stranded while RNA is single-stranded. 3. DNA uses pyrimidine thymine while RNA uses uracil.

3. What are the three classes of membrane proteins? How are each likely to function?

1. Transmembrane proteins: most likely to serve as channels or receptors. 2. Embedded proteins: most likely to have catalytic activity linked to nearby enzymes. 3. Membrane-associated (peripheral) proteins: most likely to be involved in signaling or are recognition molecules on the extracellular surface.

Khan Biotechnology Question 1

A

Von Gierke disease is an autosomal recessive disorder characterized by a deficiency in the enzyme Glucose-6-phosphatase. Which of the following would most likely be observed in an individual with Von Gierke disease? Choose 1 answer: A. Increased glycogen in the liver B. Decreased post-meal glycolysis C. Decreased glucose tolerance D. Increased gluconeogenesis

A Glucose-6-phosphatase hydrolyzes glucose-6-phosphate into a phosphate molecule plus glucose, which can then be used by cells to generate energy. Hint #22 / 5 Without this enzyme, gluconeogenesis cannot be completed. Hint #33 / 5 Glycolysis would not be affected, and glucose tolerance would be expected to be normal. Hint #44 / 5 Glycogenolysis produces glucose-6-phosphate from glycogen stores in the body during times of fasting, or when blood glucose is low. However, without glucose-6-phosphatase, glucose cannot be made from glucose-6-phosphate. Therefore, the breakdown of glycogen is impaired in these individuals. Hint #55 / 5 Increase glycogen in the liver would be expected in an individual with Von Gierke disease.

What are the names and functions of the four fat-soluble vitamins? Name Function

A (carotene) - As retinal: vision; as retinoic acid: epithelial development D (cholecalciferol) - As calcitriol: calcium and phosphate regulation E (tocopherols) - Antioxidants, using aromatic ring K (Phylloquinone and menaquinones) - Posttranslational modification of prothrombin, addition of calcium-binding sites on many proteins

How many carbons are in a diterpene?

A diterpene has 20 carbon atoms in its backbone. One terpene units is made from two isoprene units, each of which has five carbons.

2. What are the base-pairing rules according to the Watson-Crick model?

A pairs with T (in DNA) or U (in RNA), using two hydrogen bonds. C pairs with G, using 3 hydrogen bonds.

What is the difference between a steroid and a steroid hormone?

A steroid is defined by its structure: it includes three cyclohexane rings and a cyclopentane ring. A steroid hormone is a molecule within this class that also functions as a hormone, meaning that it travels in the bloodstream, is active at low concentrations, has high-affinity receptors, and affects gene expression and metabolism.

9. Which of the following is true of diffusion and osmosis? A. Diffusion and osmosis rely on the electrochemical gradient of only the compound of interest. B. Diffusion and osmosis rely on the electrochemical gradient of all compounds in a cell. C. Diffusion and osmosis will proceed in the same direction if there is only one solute. D. Diffusion and osmosis cannot occur simultaneously.

A. The movement of any solute or water by diffusion or osmosis is dependent only on the concentration gradient of that molecule and on membrane permeability.

The effect of allolactose is: Choose 1 answer: (Choice A) A Dissociation of RNA polymerase from DNA, terminating transcription of genes regulated by the lac operon (Choice B) B Enhancement of binding of RNA polymerase to DNA, increasing gene transcription (Choice C) C Conformational change in the repressor protein, inducing the lac operon (Choice D) D Repression of the lac operon when glucose is present

Allolactose is not involved in the relationship of RNA polymerase and DNA. Hint #22 / 3 Nor does allolactose deactivate the lac operon when glucose is present. Hint #33 / 3 Allolactose engenders a conformational change in the repressor protein, which causes this protein to disassociate from the lac operon, allowing for transcription. C

4. What is alternative splicing, and what does it accomplish?

Alternative splicing is the ability of some genes to use various combinations of exons to create multiple proteins from one hnRNA transcript. This increases protein diversity and allows a species to maximize the number of proteins it can create from a limited number of genes.

Which of the two forms of starch is more soluble in solution. Why?

Amylopectin is more soluble in solution than amylose because of its branched structure. The highly-branched structure of amylopectin decreases intermolecular bonding between polysaccharide polymers and increases interaction with the surrounding solution.

2. What is the relationship between osmotic pressure and the direction of osmosis through a semipermeable membrane?

As osmotic pressure increases, more water will tend to flow into the compartment to decrease solute concentration. Osmotic pressure is often considered a "sucking" pressure because water will move toward the compartment with the highest osmotic pressure.

7. Which of the following enzymes cleaves polysaccharide chains and yields maltose exclusively? A. alpha-amylase B. beta-amylase C. Debranching enzyme D. Glycogen phosphorylase

B Beta-Amylase cleaves amylose at the non-reducing end of the polymer to yield maltose exclusively, while alpha-amylase, A, cleaves amylose anywhere along the chain to yield short polysaccharides, maltose, and glucose. Debranching enzyme, C, removes oligosaccharides from a branch in glycogen or starches, while glycogen phosphorylase (D), yields glucose 1-phosphate.

How many total molecules of ATP are synthesized from ADP via glycolysis of a single molecule of glucose? Choose 1 answer: A 38 B 4 C 2 D 36

B Glycolysis is only the first step of a larger process known as cellular respiration. Hint #22 / 4 Glycolysis consumes ATP as well as producing it, though the question asks how many molecules are generated, not the "net" number generated. Hint #33 / 4 Two molecules of ATP are consumed in the "investment phase" and four molecules of ATP are generated in the "payoff phase". Hint #44 / 4 Four total molecules of ATP are generated via glycolysis of a single molecule of glucose.

Very-long chain fatty acids (VLCFAs) are unique among lipids in that they must be degraded within peroxisomes. Disorders such as adrenoleukodystrophy prevent this peroxisomal degradation, and result in an accumulation of VLCFAs throughout the body. Similarly, long-chain fatty acids (LCFAs) are known to accumulate in the setting of a deficiency of the enzyme carnitine acyltransferase (CAT). If CAT were restored to normal function, where within the cell would LCFAs be metabolized? Choose 1 answer: A Nucleus B Mitochondria C Peroxisome D Lysosome

B No fatty acid degradation occurs in the nucleus. Hint #22 / 4 Lysosomes are more of a 'trash can' for the cell, not a metabolic organelle. Hint #33 / 4 Carnitine is required for shuttling LCFAs into the organelle where they are oxidized to produce ATP and acetyl coA. Hint #44 / 4 LCFAs are degraded and metabolized in the mitochondria.

ATP is generated in the cell by extracting energy from the foods we eat. Different food sources, such as carbohydrates and dietary lipids, are broken down in different ways to achieve this goal, via numerous different biochemical pathways which are localized to discrete compartments of the cell. Where within the cell is the majority of ATP produced from the complete metabolism of carbohydrates and lipids, respectively? Choose 1 answer: A. Most of the ATP from the metabolism of carbohydrates is produced in the mitochondria, while most of the ATP from the metabolism of lipids is produced in the cytosol (Choice B) B Most of the ATP from the metabolism of carbohydrates is produced in the mitochondria, and most of the ATP from the metabolism of lipids is produced in the mitochondria (Choice C) C Most of the ATP from the metabolism of carbohydrates is produced in the cytosol, while most of the ATP from the metabolism of lipids is produced in the mitochondria (Choice D) D Most of the ATP from the metabolism of carbohydrates is produced in the cytosol, and most of the ATP from the metabolism of lipids is produced in the cytosol

B. Glycolysis is the pathway by which glucose is broken down into pyruvate. This occurs in the cytoplasm, but only generates a net of two ATP molecules. Hint #22 / 5 Lipids, as fatty acids, are metabolized in the mitochondria via beta oxidation to form acetyl-CoA. The products of glycolysis and beta oxidation both enter the TCA cycle in the matrix of the mitochondria, where NADH and FADH2 are generated. Hint #44 / 5 Mitochondrial NADH and FADH2 enter the electron transport chain and produce ATP within the mitochondria. Hint #55 / 5 Most of the ATP from the metabolism of carbohydrates is produced in the mitochondria, and most of the ATP from the metabolism of lipids is produced in the mitochondria.

13. Which of the following is true about cholesterol? A. Cholesterol always increases membrane fluidity in cells. B. Cholesterol is a steroid hormone precursor. C. Cholesterol is a precursor for vitamin A, which is produced in the skin. D. Cholesterol interacts only with the hydrophobic tails of phospholipids.

B. Cholesterol is a steroid hormone precursor that has variable effects on membrane fluidity depending on temperature, eliminating A. It interacts with both hydrophobic tails and the hydrophilic heads of membrane lipids, nullifying D. It is also a precursor for vitamin D (not vitamin A), which can be produced in the skin in a UV-driven reaction, eliminating C.

11. Which of the following is true about glycerophospholipids? A. Glycerophospholipids can sometimes be sphingolipids, depending on the bonds in their head groups. B. Glycerophospholipids are merely a subset of phospholipids. C. Glycerophospholipids are used in the ABO blood typing system. D. Glycerophospholipids have one glycerol, one polar head group, and one fatty acid tail.

B. Glycerophospholipids are a subset of phospholipids, as are sphingomyelins. Glycerophospholipids are never sphingolipids because they contain a glycerol backbone (rather than sphingosine or a sphingoid backbone), eliminating A. Sphingolipids are used in the ABO blood typing system, eliminating C. Glycerophospholipids have a polar head group, glycerol, and two fatty acid tails, not one, as in D.

Imagine that samples of each of the following carbohydrates were given a +1 charge through bombardment with electrons, and then put through a mass spectrometer. Which carbohydrate molecule would you predict to have the smallest measured deflection? A. Glucose B. Sucrose C. Mannose D. Galactose

B. Since all the carbohydrates are stipulated to have a +1 charge, the heaviest carbohydrate will deflect the least in the mass spectrometer. Hint #22 / 4 Mannose and galactose are epimers of glucose: they have the same molecular formula. Hint #33 / 4 Sucrose is a disaccharide of fructose and glucose. Hint #44 / 4 Sucrose would have the smallest measured deflection in the imagined experiment.

The antibiotic doxycycline is known to bind and inhibit a particular ribosomal subunit to inhibit bacterial proliferation. How might this affect the 80S ribosome in human cells versus the 70S subunit in prokaryotes and why? Choose 1 answer: (Choice A) A Doxycycline would not affect human ribosomes, because human ribosomes are made up of 30S and 50S subunits while prokaryotic ribosomes are made up of 30S and 40S subunits (Choice B) B Doxycycline would inhibit human ribosome function by binding the 40S subunit, because both humans and prokaryotes share the 40S subunit (Choice C) C Doxycycline would not affect human ribosomes, because human ribosomes are made up of 60S and 40S subunits while prokaryotic ribosomes are made up of 50S and 30S subunits (Choice D) D Doxycycline would inhibit human ribosome function by binding the 30S subunit, because both humans and prokaryotes share the 30S subunit

C Eukaryotes have a 60S and a 40S subunit. Hint #22 / 5 Prokaryotes have a 50S and a 30S subunit. Hint #33 / 5 The ribosome is not named by the sum of the subunits; prokaryotes have a 70S ribosome and eukaryotes have an 80S ribosome. Hint #44 / 5 Side note: Doxycycline acts on the 30S subunit. Hint #55 / 5 Doxycycline would not affect human ribosomes, because human ribosomes are made up of 60S and 40S subunits while prokaryotic ribosomes are made up of 50S and 30S subunits.

8. An investigator is measuring the activity of various enzymes involved in reactions of intermediary metabolism. One of the enzymes has greatly decreased activity compared to reference values. The buffer of the assay contains citrate. Which of the following enzymes will most likely be directly affected using citrate? A. Fructose-2,6-bisphosphatase B. Isocitrate dehydrogenase C. Phosphofructokinase-1 D. Pyruvate carboxylase

C. Citrate is produced by citrate synthase from acetyl-CoA and oxaloacetate. This reaction takes place in the mitochondria. When the citric acid cycle slows down, citrate accumulates. In the cytosol, it acts as a negative allosteric regulator of phosphofructokinase-1, the enzyme that catalyzes the rate-limiting step of glycolysis.

15. The cyclic forms of monosaccharides are: I. Hemiacetals II. Hemiketals. III. Acetals. A. I only B. III only C. I and II only D. I, II, and III

C. Monosaccharides can exist as hemiacetals or hemiketals, depending on whether they are aldoses or ketoses. When a monosaccharide is in its cyclic form, the anomeric carbon is attached to the oxygen in the ring and a hydroxyl group. Hence, it is only hemiacetal or hemiketals because an acetal or ketal would require the -OH group to be converted to another -OR group.

11. Why might uracil be excluded from DNA but NOT RNA? A. Uracil is much more difficult to synthesize than thymine. B. Uracil binds adenine too strongly for replication. C. Cytosine degradation results in uracil. D. Uracil is used as a DNA synthesis activator.

C. One common DNA mutation is the transition from cytosine to uracil in the presence of heat. DNA repair enzymes recognize uracil and correct this error by excising the base and inserting cytosine. RNA exists only transiently in the cell, such that cytosine degradation is insignificant. Were uracil to be used in DNA under normal circumstances, it would be impossible to tell if a base should be uracil or if it is a damaged cytosine nucleotide.

Which procedure is least effective for detecting a specific DNA sequence in a sample? Choose 1 answer: (Choice A) A Southern Blot (Choice B) B Polymerase Chain Reaction (Choice C) C Gel Electrophoresis (Choice D) D Microarray

C. Polymerase Chain Reaction uses specially constructed primers to amplify a specific DNA sequence. Thus, you can tell if the target sequence was in the sample based on if the reaction produces more nucleic acid. A Southern blot is one of the most efficient methods for detecting the presence of a specific DNA sequence in a sample. It works via hybridization of the target sequence to a probe. If the probe latches on, the target is there. Gel Electrophoresis separates fragments based on size, and so is not nearly as specific as the other techniques mentioned above.

8. When the following straight-chain Fischer projection is converted to a chair or ring conformation, its structure will be: (see structures in carbohydrate structure and function)

C. Start by drawing out the Haworth projection. Recall that all the groups on the right in the Fischer projection will go on the bottom of the Haworth projection, and all the groups on the left will go on the top. Next, draw the chair structure, with the oxygen in the back right corner. Label the carbons in the ring 1 through 5, starting from the oxygen and moving clockwise around the ring. Now, draw the lines for all the axial substituents, alternating above and below the ring. Remember to start on the anomeric C-1 carbon, where the axial substituent points down. Now start filling in the substituents. The substituent can be in either position on the anomeric carbon, so skip that one for now. The -OH groups on C-2 and C-4 should point downwards while the -OH group on C-3 should point upwards; C, the beta-anomer of D-glucose, is the only one that matches.

4. Val-tRNAVal is the tRNA that carries valine to the ribosome during translation. Which of the following sequences gives an appropriate anticodon for this tRNA? (Note: Refer to Figure 7.5 for a genetic code table). A. CAU B. AUC C. UAC D. GUG

C. There are four different codons for valine: GUU, GUC, GUA, and GUG. Through base-pairing, we can determine that the proper anticodon must end with "AC." Remember that the codon and anticodon are antiparallel to each other, and that nucleic acids are always written 5' 3' on the MCAT. Therefore, we are looking for an answer that ends with "AC" (rather than starting with "CA").

3. What are the three main sites of regulation within the citric acid cycle? What molecules inhibits and activate the three main checkpoints?

Checkpoints Inhibitors Activators Citrate synthase Inhibitors: ATP, NADH, citrate, succinyl-CoA Activators: None Isocitrate dehydrogenase Inhibitors: ATP, NADH Activators: ADP, NAD+ Alpha-ketoglutarate complex Inhibitors: ATP, NADH, succinyl-CoA Activators: ADP, Ca2+

2. Compare heterochromatin and euchromatin based on the following characteristics: Characteristic Density of chromatin packing Appearance under light microscopy Transcription activity

Characteristic Density of chromatin packing Heterochromatin: Dense Euchromatin: Not dense (uncondensed) Appearance under light microscopy Heterochromatin: Dark Euchromatin: Light Transcription activity Heterochromatin: Silent Euchromatin: Active

Which change in solution composition would cause a protein to elute from a hydrophobic interaction column? A Decreasing pH B Increasing pH C Decreasing salt concentration D Increasing salt concentration

Choice C is the correct answer. Hydrophobic interaction chromatography (HIC) relies on high salt concentrations to enhance or strengthen hydrophobic interactions. Therefore, decreasing the salt concentration weakens these interactions, causing the protein to dissociate from the column. There is not a significant role for pH in HIC, making A and B incorrect. There could be a small effect if charges were to be neutralized but this is not the primary effect. Answer D is not correct since this would strengthen interactions with the column, not weaken them.

2. How does cholesterol play a role in the fluidity and stability of the plasma membrane?

Cholesterol moderates membrane fluidity by interfering with the crystal structure of the cell membrane and occupying space between phospholipid molecules at low temperatures, and by restricting excessive movement of phospholipids. Cholesterol also provides stability by cross-linking adjacent phospholipids through interactions at the polar head group and hydrophobic interactions at the nearby fatty acid tail.

1. What is the purpose of all the reactions that collectively make up the citric acid cycle?

Complete oxidation of carbons in intermediates to CO2 so that reduction reactions can be coupled with CO2 formation, thus forming energy carriers such as NADH and FADH2 for the electron transport chain.

An organic molecule has been isolated from the nucleus of a human cell. Nuclear magnetic resonance studies reveal the core structure of the molecule to be a six-membered ring containing carbon and nitrogen. The ring also expresses an amine group and a keto group, and is attached to a five carbon sugar moiety. No phosphate groups are noted. What is the most likely identity of the molecule? A. Cytosine B. Guanosine C. Guanine D. Cytidine

D First, realize that this question is asking you to identify a likely molecular component of the nucleus, DNA. DNA is largely made up of the nucleobases Adenine, Guanine, Cytosine, and Thymine. Hint #22 / 5 When these bases are attached to a ribose moiety, the five carbon sugar described in the question, they are called "nucleosides", and are named as Adenosine, Guanosine, Cytidine, and Thymidine. Hint #33 / 5 There are two main categories of nucleobases. Pyrimidines are made of a six membered heterocyclic ring of carbon and nitrogen with various substitutions on it's four carbon atoms. Purines are made of a pyrimidine plus an imidazole ring made of two nitrogens and a carbon atom. Hint #44 / 5 Adenosine and Guanosine are nucleosides made from purines while Thymidine and Cytidine are made from pyrimidines. Hint #55 / 5 The most likely identity of the molecule in question is Cytidine.

10. Which two polysaccharides share all their glycosidic linkage types in common? A. Cellulose and amylopectin B. Amylose and glycogen C. Amylose and cellulose D. Glycogen and amylopectin

D Glycogen and amylopectin are the only polysaccharide forms that demonstrate branching structure, making them most similar in terms of linkage. Both glycogen and amylopectin use alpha-1,4 and alpha-1,6 linkages. Cellulose uses B-1,4 linkages and amylose does not contain alpha-1,6 linkages.

At what point during normal DNA replication is genetic material lost from the telomeres? Choose 1 answer: A Enzymatic action of telomerase B Attachment of DNA polymerase to the leading strand C "Unzipping" by DNA helicase D Joining of adjacent Okazaki fragments

D Telomeres are the 'caps' at either end of a chromatid, which protects the DNA from deterioration. It has been theorized that telomere length is inversely correlated with the aging process (shorter telomeres = aging). Hint #22 / 5 The shortening of telomeres is due to the 'end replication problem'. This problem arises from the fact that DNA polymerase synthesizes DNA only in the 5' to 3' direction. One strand (the leading strand) is read without interruption, but the lagging strand must be synthesized in a discontinuous fashion. Hint #33 / 5 Okazaki fragments are pieces of newly synthesized DNA along the lagging DNA strand, separated by RNA primers. Hint #44 / 5 The final RNA primer at the terminal end of the new strand cannot be converted to DNA by DNA polymerase and joined to the strand by DNA ligase, because there is nowhere for these enzymes to bind. As such, this last primer is degraded, and thus each successive DNA copy is slightly shorter than the last. Hint #55 / 5 Genetic material lost from the telomeres during the joining of Okazaki fragments.

Draw all the possible D-stereoisomers of glucose in Fisher projection form.

D-allose, D-altrose, D-glucose, D-mannose, D-gulose, D-idose, D-galactose, D-talose

7. A membrane receptor is most likely to be a(n): A. embedded protein with catalytic activity B. transmembrane protein with sequestration activity C. membrane-associated protein with sequestration activity. D. transmembrane protein with catalytic activity

D. Membrane receptors must have both an extracellular and intracellular domain; therefore, they are considered transmembrane proteins. To initiate a second messenger cascade, they typically display enzymatic activity, although some may act strictly as channels.

3. During DNA sequencing, why does the DNA polymer stop growing once a dideoxyribonucleotide is added?

Dideoxyribonucleotides lack the 3' -OH group that is required for DNA strand elongation. Thus, once a dideoxyribonucleotide is added to a growing DNA molecule, no more nucleotides can be added because dideoxyribonucleotides have no 3' -OH group with which to form a bond.

1. What is the primary thermodynamic factor responsible for passive transport?

Entropy

1. For each of the enzymes listed below, list the function of the enzyme and if they are found in prokaryotes, eukaryotes, or both. Enzyme Helicase Single-stranded DNA-binding protein Primase DNA polymerase III DNA polymerase alpha DNA polymerase I RNase H DNA ligase DNA topoisomerases

Enzyme Prokaryotes/Eukaryotes/Both Function Helicase Both Unwinds DNA double helix Single-stranded DNA-binding protein Both Prevents reannealing of DNA double helix during replication Primase Both Places ~10-nucleotide RNA primer to begin DNA replication DNA polymerase III Prokaryotes Adds nucleotides to growing daughter strand DNA polymerase alpha Eukaryotes Adds nucleotides to growing daughter strand DNA polymerase I Prokaryotes Fills in gaps left behind after RNA primer excision RNase H Eukaryotes Excises RNA primer DNA ligase Both Joins DNA strands (especially between Okazaki fragments) DNA topoisomerases Both Reduces torsional strain from positive supercoils by introducing nicks in DNA strand

Explain the difference between esterification and glycoside formation.

Esterification is the reaction by which a hydroxyl group reacts with either a carboxylic acid or a carboxylic acid derivative to form an ester. Glycoside formation refers to the reaction between an alcohol and a hemiacetal (or hemiketal) group on a sugar to yield an alkoxy group.

9. During mRNA processing, mRNA is modified with a 5' cap and a 3' poly-adenine tail. While the poly-A tail can have as many as 250 nucleotides, the 5' cap only has 1 nucleotide. What difference(s) could this difference? A. The 5' cap functions exclusively to aid in nuclear export, while the poly-a tail's primary function is to bind to the ribosome. B. Polyadenylation doesn't begin until elongation is complete, while the 5' cap is added during elongation. C. The 5' cap is unable to protect against RNA degradation and has other functions. D. Nucleotides at the 5' end of nucleic acid are more efficient than ones at the 3' end.

For this question, consider the functions of the 5' cap and the 3' poly-A tail in mature mRNA. Both protect against degradation and help with nuclear export. The guanine cap, however, is added during transcription, while the adenine tail is added after.

Regarding glycogen and amylopectin, which of these two polymers should experience a higher rate of enzyme activity from enzymes that cleave side branches? Why?

Glycogen has a higher rate of enzymatic branch cleavage because it contains significantly more branching than amylopectin.

Which statement concerning the glycolytic and gluconeogenic pathways is correct? Choose 1 answer: A Gluconeogenesis is anabolic and glycolysis is catabolic B Both are catabolic C Gluconeogenesis is catabolic and glycolysis is anabolic D Both are anabolic

Glycolysis is the breakdown of glucose; gluconeogenesis is the synthesis of glucose. Hint #22 / 4 Glycolysis yields a net gain of ATP while gluconeogenesis yields a net loss. Hint #33 / 4 Catabolism can be thought of as the breakdown of larger molecules into smaller ones, while anabolism can be thought of as the opposite. Hint #44 / 4 Gluconeogenesis is anabolic and glycolysis is catabolic.

3. What property of telomeres and centromeres allows them to stay tightly raveled, even when the rest of DNA is uncondensed?

High GC-content increases hydrogen bonding, making the association between DNA strands very strong at telomeres and centromeres.

2. By what histone and DNA modifications can genes be silenced in eukaryotic cells? Would these processes increase the proportion of heterochromatin or euchromatin?

Histone deacetylation and DNA methylation will both downregulate the transcription of a gene. These processes allow the relevant DNA to be clumped more tightly, increasing the proportion of heterochromatin.

2. What enzyme catalyzes the rate-limiting step of the citric acid cycle?

Isocitrate dehydrogenase

. From a metabolic standpoint, does it make sense for carbohydrates to get oxidized or reduced? What is the purpose of this process?

It makes sense for carbohydrates to become oxidized while reducing other groups. This is the case because aerobic metabolism requires reduced forms of electron carrier to facilitate processes such as oxidative phosphorylation. Because carbohydrates are a primary energy source, they are oxidized.

What would happen if an amphipathic molecule were placed in a nonpolar solvent rather than an aqueous solution?

It would form a micelle so that the fatty acid/hydrophobic tails faced the nonpolar solvent.

2. List the following membrane components in order from plentiful to least plentiful: carbohydrates, lipids, proteins, nucleic acids.

Lipids, including phospholipids, cholesterol, and others, are most plentiful; proteins, including transmembrane proteins (channels and receptors), membrane-associated proteins, and embedded proteins, are next most plentiful; carbohydrates, including the glycoprotein coat and signaling molecules, are next; nucleic acids are essentially absent.

2. What other molecules can be used to make acetyl-CoA, and how does the body perform this conversion for each?

Molecule Mechanism of Conversion to Acetyl-CoA Fatty acids Shuttle acyl group from cytosolic Co-A-SH to mitochondrial CoA-SH via carnitine; then undergo beta-oxidation. Ketogenic amino acids Transaminate to lose nitrogen; convert carbon skeleton into ketone body, which can be converted into acetyl-CoA. Ketones Reverse of ketone body formation Alcohol Alcohol dehydrogenase and acetaldehyde dehydrogenase convert alcohol into acetyl-CoA

1. What is the difference between a nucleoside and a nucleotide?

Nucleosides contain a five-carbon sugar (pentose) and nitrogenous base. Nucleotides are composed of a nucleoside plus one to three phosphate groups.

4. What is the key structural difference in the types of lesions corrected by nucleotide excision repair vs. those corrected by base excision repair?

Nucleotide excision repair corrects lesions that are large enough to distort the double helix; base excision repair corrects lesions that are small enough not to distort the double helix.

1. What is the difference between an oncogene and a tumor suppressor gene?

Oncogenes (or, more properly, proto-oncogenes) code for cell cycle-promoting proteins; when mutated, a proto-oncogene becomes an oncogene, promoting rapid cell cycling. Tumor suppressor genes code for repair or cell cycle-inhibiting proteins; when mutated, the cell cycle can proceed unchecked. Oncogenes are like stepping on the gas pedal, mutated tumor suppressor genes are like cutting the brakes.

3. Compare the two types of active transport. What is the difference between symport and antiport?

Primary active transport uses ATP as an energy source for the movement of molecules against their concentration gradient, while secondary active transport uses an electrochemical gradient to power the transport. Symport moves both particles in secondary active transport across the membrane in the same direction, while antiport moves particles across the cell membrane in opposite directions.

NSAIDs block prostaglandin production to reduce pain and inflammation. What do prostaglandin do to bring about these symptoms?

Prostaglandins regulate the synthesis of cAMP, which is involved in many pathways, including those that drive pain and inflammation.

1. What is the overall reaction of the pyruvate dehydrogenase complex?

Pyruvate + CoA + NAD+ --> Acetyl-CoA + CO2 + NADH + H+

2. When starting transcription, where does RNA polymerase bind?

RNA polymerase II binds to the TATA box, which is located within the promoter region of a relevant gene, at about -25.

2. From 5' to 3', what are the component of the operon, and what are their roles?

Regulator gene Transcribed to form repressor protein Promoter site Site of RNA polymerase binding (like promoters in eukaryotes) Operator site Binding site for repressor protein Structural gene The gene of interest; its transcription is dependent on the repressor being absent from the operator site

3. For each of the repair mechanisms below, in which phase of the cell cycle does the repair mechanism function? What are the key enzymes or genes specifically associated with each mechanism? Repair Mechanism DNA polymerase (proofreading) Mismatch repair Nucleotide excision repair Base excision repair

Repair Mechanism Phase of Cell Cycle Key Enzymes/Genes DNA polymerase (proofreading) S DNA polymerase Mismatch repair G2 MSH2, MLH1 (MutS and MutL in prokaryotes) Nucleotide excision repair G1, G2 Excision endonuclease Base excision repair G1, G2 AP endonuclease, glycosylase

1. In an enhancer, what are the differences between signal molecules, transcription factors, and response elements?

Signal molecules include steroid hormones and second messengers, which bind to their receptors in the nucleus. These receptors are transcription factors that use their DNA-binding domain to attach to a sequence in DNA called a response element. Once bonded to the response element, these transcription factors can then promote increased expression of the relevant gene.

4. Why does soap appear to dissolve in water, and how is this fact important to cleaning?

Soap appears to dissolve in water because amphipathic free fatty acid salts form micelles, with hydrophobic fatty acid tails toward the center and carboxylate groups facing outward toward the water. Fat-soluble particles can then dissolve inside micelles in the soap-water solution and wash away. Water-soluble compounds can freely dissolve in water.

5. For each of the mutations listed below, what changes in DNA sequence are observed, and what effect do they have on the encoded peptide? Silent (degenerate) Missense Nonsense Frameshift

Type of Mutation Change in DNA Sequence Effect on Encoded Protein Silent (degenerate) Substitution of bases in the wobble position, introns, or noncoding DNA No change observed Missense Substitution of one base, creating an mRNA codon that matches a different amino acid One amino acid is changed in the protein; variable effects on function depending on specific change Nonsense Substitution of one base, creating a stop codon Early truncation of protein; variable effects on function, but usually more severe than missense mutations Frameshift Insertion or deletion of bases, creating a shift in the reading frame of the mRNA Change in most amino acids after the site of insertion or deletion; usually the most severe of the types listed here

3. What is the function of a telomere?

Telomeres are the ends of eukaryotic chromosomes and contain repetitive sequences of noncoding DNA. These protect the chromosome from losing important genes from the incomplete replication of the 5' end of the DNA strand.

4. How does the aromaticity of purines and pyrimidines underscore their genetic function?

The aromaticity of nucleic acids makes these compounds very stable and unreactive. Stability is important for storing genetic information and avoiding spontaneous mutations.

What is the difference between a sphingolipid that is also a phospholipid and one that is NOT?

The difference is the bond between the sphingosine backbone and the head group. When this is a phosphodiester bond, it's a phospholipid (note the phospho-prefixes). Nonphospholipid sphingolipids include glycolipids, which contain a glycosidic linkage to sugar.

3. What bonds are broken during saponification?

The ester bonds of triacylglycerols are broken to form a glycerol molecule and the salts of fatty acids (soap).

1. What are the five histone proteins in eukaryotic cells? Which one is not part of the histone core around which DNA wraps to form chromatin?

The five histone proteins are H1, H2A, H2B, H3, and H4. H1 is not part of the histone core.

1. How does the human body store spare energy? Why doesn't the human body store most energy as sugar?

The human body stores energy as glycogen and triacylglycerols. Triacylglycerols are preferred because their carbons are more reduced, resulting in a larger amount of energy per unit weight. In addition, due to their hydrophobic nature, triacylglycerols do not need to cary extra weight from hydration.

3. What distinguishes the inner mitochondrial membrane from other biological membranes? What is the pH gradient between the cytoplasm and the intermembrane space?

The inner mitochondrial membrane lacks cholesterol, which differentiates it from most other biological membranes. There is no pH gradient between the cytoplasm and the intermembrane space because the outer mitochondrial membrane has such high permeability to biomolecules (the proton-motive force of the mitochondria is across the inner mitochondrial membrane, not the outer mitochondrial membrane).

In a bacteria possessing the lac operon, which of the following occurs when glucose is low and lactose is abundant? Choose 1 answer: A Beta-galactosidase acetylases the operon B Lactose permease cleaves lactose to glucose and galactose C Lactose metabolism is increased by lactose binding to the operon D Transport of lactose into the cell is enhanced

The lac operon contains genes necessary to enhance lactose metabolism. Hint #22 / 3 Cleavage of lactose is carried out by the beta-galactosidase encoded for by lacZ, not by lactose permease. Hint #33 / 3 When glucose is low but lactose is readily available, the lac operon is activated, enhancing transport of lactose into the cell as well as lactose utilization. D

2. Between the leading strand and lagging strand, which is more prone to mutations? Why?

The lagging strand is more prone to mutations because it must constantly start and stop the process of DNA replication. Additionally, it contains many more RNA primers, all of which must be removed and filled in with DNA.

1. How is the resting membrane potential maintained?

The membrane potential, which results from a difference in the number of positive and negative charges on either side of the membrane, is maintained primarily by the sodium-potassium pump, which moves three sodium ions out of the cell for every two potassium ions pumped in, and to a minor extent by leak channels that allow the passive transport of ions.

2. How does DNA polymerase recognize which strand is the template strand once the daughter strand is synthesized?

The parent strand is more heavily methylated, whereas the daughter strand is barely methylated at all. This allows DNA polymerase to distinguish between the two strands during proofreading.

5. If a strand of RNA contained 15% cytosine, 15% adenine, 35% guanine, and 35% uracil, would this violate Chargaff's rules? Why or why not?

This does not violate Chargaff's rules. RNA is single-stranded, and thus the complementarity seen in DNA does not hold true. For single-stranded RNA, %C does not necessarily equal %G; %A does not necessarily equal %U.

20. After running a DNA gel and identifying particular DNA fragments using protein probes, the scientist wishes to remove the proteins and just have the nucleic acid sequences present. He runs the euchromatin DNA through a special ion-exchange chromatogram via a carboxymethyl-cellulose (anionic) lined column. What will most likely NOT be removed first via this step? A. Nuclear factor I (NFI) transcription factor B. CREB C. TATA box binding protein (TBP) D. Histone H3

This question has a lot of info so be sure to capture all of the important pieces. We have a DNA molecule that is mostly euchromatin with protein attached and we want to remove the protein by putting it into a column with negatively-charged residues. Thus, positively charged molecules will bind the column and be removed. Theoretically, all DNA binding proteins are somewhat positively charged b/c they bind to the negatively charged DNA backbone. However, euchromatin is a transcriptionally-active state and histone proteins should be acetylated to allow for a more relaxed structure. Histone acetylation removes the positive charges on the lysine residues in the histones and thus they would not bind as well to the anionic column. The other choices are proteins that would be avidly bound to DNA b/c of their positive charges. D

5. A microbiologist sets out to discover whether archaeal messenger RNA is monocistronic or polycistronic. Which of the following pieces of evidence most clearly supports only one of the two conclusions? A. Cytoplasmic mRNA analysis shows multiple tRNA molecules attached. B. Electron micrographs of translation shows a ribosomal RNA protein complex being formed. C. Multiple AUG codons are found in the mRNA sequence. D. Multiple UAG, UAA, or UGA present in the reading frame of the mRNA.

To answer this "supporting evidence" question, identify a choice which supports only one of the two possibilities: monocistronic or polycistronic. A, B and C are true for both possibilities. D, however, would be true of only polycistronic mRNA, which is an mRNA containing the code for multiple proteins and would thus have multiple different termination sequences.

21. HpaII is a restriction enzyme used for analyzing DNA methylation. It cleaves 5'-CCGG-3' sites but will not if there is methylation of the cytosine residues. A sample of genomic DNA from different organisms is incubated with HpaII and analyzed via agarose gel electrophoresis. Which of the following choices best explains the patterns? See table A. Drosophila and E. coli contain methylated DNA initially but DNA methylases cleave methyl groups off early in their lifespans. B. The rabbit DNA contains silent genes that do not need to be turned on. C. About 75% of rabbit DNA is methylated D. Drosophila and E. coli DNA is heavily methylated and thus very little enzyme digestion occurs.

To answer this question first examine the table. Rabbit DNA contains some small fragments but has a lot more long fragments. This means that most of it is not being digested by HpaII due to methylation. On the other hand, most of the other two species' DNA is digested. Thus, their DNA is mostly un-methylated. The consequence of methylation is typically the silencing of genes that don't need to be turned on. B.

1. A student devises a probe to find the source of the rRNA for a particular cell. He discovers that, instead of finding one signal, as he previously thought, there are 4 separate signals that fluorescently light up. What best explains this observation? A. The cell the student was examining does not contain any rRNA. B. The student incorrectly devised the nuclear probe and should have used fewer nucleotides. C. The cell is metabolically highly active and thus contains several nucleoli. D. The nucleic acid he found on fluorescence probing is used to synthesize not only rRNA but also tRNA and mRNA.

To answer this question, consider where in the cell rRNA is produced: the nucleolus. Cells can have multiple nucleoli depending on the activity so C is correct. The probe may have been faulty, but that would likely yield more erroneous results and the above explanation is more likely. An rRNA gene-specific probe should not bind to genes coding for tRNA and mRNA, as D says.

4. What is the difference between a transgenic and a knockout mouse?

Transgenic mice have a gene introduced into their germ line or embryonic stem cells to look at the effects of that gene; they are therefore best suited for studying the effects of dominant alleles. Knockout mice are those in which a gene of interest has been removed, rather than added.

2. Describe the structure and function of triacylglycerols.

Triacylglycerols, also called triglycerides, are composed of a glycerol backbone esterified to three fatty acids. They are used for energy storage.

Name the three main types of sphingolipids and their characteristics. Type Phospholipid or Glycolipid? Functional Group(s)

Type Phospholipid or Glycolipid? Functional Group(s) Type: Sphingomyelin Phospholipid Functional Groups: Phosphatidylcholine or phosphatidylethanolamine Type: Glycosphingolipid Glycolipid Functional Groups: Sugars bonded by glycosidic linkages Type: Gangliosides Glycolipid Functional groups: Oligosaccharides with one or more N-acetylneuraminic acid (NANA or sialic acid)

15. A commonly employed technique in molecular biology is a "knockout" mutation, in which the gene is turned off essentially. Describe the result when the regulator gene for the lac operon is knocked out. A. The lac genes would be expressed efficiently only in the presence of lactose. B. The lac genes would be expressed efficiently only in the absence of lactose. C. The lac genes would never be expressed efficiently. D. The lac genes would be expressed constitutively.

With no regulator gene, no repressor protein can be made. In the absence of repressor protein, the lac gene is expressed constitutively. D.

17. What is the most likely status of the trp operon in bacteria that have been successfully cultured in a medium containing fully formed proteins and electrolytes but no free amino acids? A. The trp repressor tetramers can bind to the operator of the trp operon. B. Even in the presence of repressor, the structural genes are turned on. C. The trp operon is partially repressed due to the presence of electrolytes that short-circuit the operon. D. The trp repressor tetramers have no tryptophan to bind and dissolve in the nucleoplasm.

Without any growth media or nutrition, the bacteria have to rely on any and all fuel sources. Protein will be degraded to amino acids and the high levels of tryptophan will shut down the biosynthesis trp operon. A.

4. What is wobble, and what role does it serve?

Wobble refers to the fact that the third base in a codon often plays no role in determining which amino acid is translated from that codon. For example, any codon starting with "CC" codes for proline, regardless of which base is in the third (wobble) position. This is protective because mutations in the wobble position will not have any effect on the protein translated from that gene.

2. The three-base sequences listed below are DNA sequences. Using Figure 7.5, which amino acid is encoded by each of these sequences, after transcription and translation? GAT ATT CGC CCA

• GAT: mRNA codon = AUC; Isoleucine (Ile) • ATT: mRNA codon = AAU: Asparagine (Asn) • CGC: mRNA codon = GCG: Alanine (Ala) • CCA: mRNA codon = UGG: Tryptophan (Trp)

4. Contrast gap junctions and tight junctions.

• Gap junctions: allow for intercellular transport of materials and do not prevent paracellular transport of materials. • Tight junctions: are not used for intercellular transport but do prevent paracellular transport. Gap junctions are discontinuous bunches around the cell, while tight junctions form bands around the cell.

1. When creating a DNA library, what are some of the advantages of genomic libraries? What about cDNA libraries?

• Genomic: Genomic libraries include all the DNA in an organism's genome, including noncoding regions. This may be useful for studying DNA in introns, centromeres, or telomeres. • cDNA: cDNA libraries only include expressed genes from a given tissue, but can be used to express recombinant proteins or to perform gene therapy.

2. What does PCR accomplish for a researcher? What about Southern blotting?

• PCR: PCR increases the number of copies of a given DNA sequence and can be used for a sample containing very few copies of the DNA sequence. • Southern blotting: Southern blotting is useful when searching for a DNA sequence because it separates DNA fragments by length and then probes for a sequence of interest.

3. What is a positive control system? What is a negative control system?

• Positive control system: require the binding of a protein to the operator site to increase transcription. • Negative control system: require the binding of a protein to the operator site to decrease transcription

1. What is the role of each eukaryotic RNA polymerase? RNA polymerase I RNA polymerase II RNA polymerase III

• RNA polymerase I: synthesizes most rRNA • RNA polymerase II: synthesizes mRNA (hnRNA) and snRNA. RNA polymerase III: synthesizes tRNA and some rRNA

3. Which mRNA is the start codon, and what amino acid does it code for? Which mRNA codons are the stop codons?

• Start codon: AUG; codes for: Methionine • Stop codons: UAA, UGA, UAG

Which components of membrane lipids contribute to their structural role in membranes? Which components contribute to function?

• Structure: Membrane lipids are amphipathic: they have hydrophilic heads and hydrophobic tails, allowing for the formation of bilayers in aqueous solution. The fatty acid tails form the bulk of the phospholipid bilayer, and play a predominantly structural role. • Function: On the other hand, the functional differences between membrane lipids are determined by the polar head group, due to its constant exposure to the exterior environment of the phospholipid bilayer (remember, this can be either the inside or outside of the cell). The degree of unsaturation of fatty acid tails can also play a small role in function.

1. What are the roles of the three main types of RNA?

• mRNA: carries information from DNA by traveling from the nucleus (where it is transcribed) to the cytoplasm (where it is translated). • tRNA: translated nucleic acids to amino acids by pairing its anticodon with mRNA codons; it is charged with an amino acid, which can be added to the growing peptide chain. • rRNA: forms much of the structural and catalytic component of the ribosome, and acts as a ribozyme to create peptide bonds between amino acids.

1. What type of operon is the trp operon? The lac operon?

• trp: Negative repressible system • lac: negative inducible system