NS 2 biology
D is correct. If the mass of the bacteria were to double every 30 min, beginning with 1 g, then the sequence of mass doubling can be represented as below in the simplest terms: 1 2 4 8 16 32 64 128 256 512 1024 This shows that the mass would be 583.8 g during the final interval represented above, which occurs after 270 min (4.5 hours) have elapsed but before 300 min (5 hours) have passed. A, B, C: In all of these choices, not enough time has elapsed for the bacterial mass to reach 583.8 g.
A sample of bacteria has a doubling time of 30 minutes. If a student begins with 1 g of the bacteria, at what time will the sample weigh 583.8 g? A. In less than an hour B. Between 1.5 and 2 hours C. Between 2.5 and 3 hours D. Between 4.5 and 5 hours
B: For ΔG to equal zero, Keq would need to be equal to 1, which is not true here.
Keq would need to be equal to 1
C: ATP is not required for reactions that are already spontaneous (ΔG < 0).
(ΔG < 0) what is not required
A: Since Keq > 1, ΔG must be negative (below zero). This relationship comes from the equation ΔG° = -RT ln (Keq).
A: Since Keq > 1, ΔG must be
B is correct. The passages states that high levels of NADPH inhibit G6PD, and Equation 1 shows that the substrate of G6PD is G6P. Because the structures of G6P and NADPH are very different, it is unlikely that that NADPH competes with G6P at the active site; thus, this is not competitive inhibition. The passage also never suggests that the inhibition of G6PD is irreversible; rather, it is likely to be dynamic based on the amount of NADPH available. Therefore, NADPH most probably binds to a site that is not the active site, which is characteristic of allosteric inhibition. A generic model of allosteric inhibition is included below.
Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD? I. Competitive II. Allosteric III. Irreversible A. I only B. II only C. II and III only D. I, II, and III
C is correct. We could do the math for this question and determine the exact mass of each of the answer choices. However, we want to work smarter, not harder, so let's see if we can determine the smallest (by mass) molecule here. By far, the smallest molecule listed is the two-amino-acid chain of alanine and leucine. The heaviest amino acid, tryptophan, weights 204 Da. Thus, two bound residues must weigh 408 Da or less. This is much smaller than the other answer choices. If you did the math, you would find that each amino acid has 2 carbon, 2 oxygen, 1 nitrogen, and 4 hydrogen atoms. Alanine has a methyl side chain (1 C, 3 H), while leucine has an isobutyl side chain (4 C, 9 H), for an additional 5 carbon and 12 hydrogen atoms and a total of 9 C, 2 O, 2 N, and 20 H. This is by far the smallest molecule presented. A: One nucleotide consists of a ribose sugar, a phosphate group, and a nitrogenous base. The phosphate group contains 1 P and 4 O atoms. The ribose consists of 3 O and 5 C atoms, while the nitrogenous base includes at least 1 N and 6 C atoms. Adding this up, we get 35 + 7(16) + 11(12) + 14 = 293. So, three attached nucleotides of any kind would put us above 800 Da, well over choice C. B: Using Figure 2, we can see that a rotigaptide molecule weights about 617 Da. D: Hexadecanoic (aka palmitic) acid has the formula C16H32O2. The carbon alone weighs 192 Da, so three hexadecanoic acid molecules would weigh over 576 Da.
Based on mass, which of the following molecules will most easily pass through a gap junction? A. An RNA sequence of 3 uracil nucleotides B. Rotigaptide C. An Ala-Leu dipeptide D. A triglyceride with 3 hexadecanoic acid molecules
B is correct. The passage discusses the role of gap junctions in conducting nerve impulses through cardiac muscle with the appropriate timing needed for prompt, regular contractions. Disrupting this flow would mean that some parts of the heart would contract before others, resulting in an abnormal heart rhythm or arrhythmia. A, D: Gap junction dysfunction would not be expected to affect either blood vessel walls or the function of valves in the heart. C: Although the total force of contraction might be reduced if not all regions of the heart muscle are contracting in an orchestrated way, the actual contractile strength of heart muscle tissue would not itself be reduced.
Based on passage information, cardiac disorders involving gap junction dysfunction would most likely manifest clinically through: A. narrowing of blood vessels. B. arrhythmia. C. weakening of cardiac muscle. D. incomplete valve closure.
B is correct. This question asks us to determine why bacterial infections of the brain are more serious than normal infections. The passage states that large molecules have a hard time getting past the blood-brain barrier. Thus, choice B is correct. A: This choice might be appealing, but there is no reason to think that the strength of bacteria influences its ability to cross the BBB. The barrier only discriminates based on the size and polarity of particles. C, D: These statements are untrue and are not supported by information in the passage
Based on the information presented in the passage, which of the following is most likely true about infections of the CNS? A. Bacterial infections of the brain are more serious because only the strongest bacteria can make it through the blood-brain barrier. B. Bacterial infections of the brain are more serious because antibiotics or other treatments have a hard time passing the blood-brain barrier. C. Bacterial infections of the brain are more serious because the brain is such an essential organ that there are not many treatments that can effectively treat the infection without serious side effects. D. Bacterial infections of the brain are stronger than normal bacterial infections.
Cardiac Contraction The rhythm of heart contractions is controlled by the sinoatrial (SA) node, which is found at the roof of the right atrium. The cells in the SA node periodically send out action potentials, much like nerve cells. Gap junctions between the cardiac muscle cells allow the action potential to propagate throughout the tissue, causing contraction. The action potential flows from the SA node into the atria, but not into the ventricles because of a layer of insulating tissue. This causes both atria to contract, pushing the blood forward into the ventricles. The atrioventricular (AV) node allows the action potential to pass through to the ventricles after the atria have contracted. At this point, the ventricles must contract to push the blood out of the heart. Since the ventricles are larger than the atria, it is a bit more difficult to get them to contract together, so the signal is sped through the bundle of His and Purkinje fibers to all the muscle cells of the ventricles. Deoxygenated blood returns to the right atrium via the superior and inferior venae cavae and the coronary sinus, which drains the coronary veins. From there, it is pumped into the right ventricle through the tricuspid valve. From the right ventricle, it goes to the pulmonary arteries through the pulmonary semilunar valves. After becoming oxygenated, it is returned to the heart via the pulmonary veins, and enters the left atrium. It is pumped through the bicuspid valve from the left atrium to the left ventricle, and then the left ventricle pushes the blood into circulation (more specifically, through the aortic semilunar valves into the ascending aorta).
Cardiac Contraction
Cell Cycle and Checkpoints The cell cycle can be divided into a resting phase (interphase) and cell division (mitosis or meiosis). Resting phase is also known as Gap 0 (G0). During this period, the cell just goes about its business; in fact, many fully-differentiated cells in the body remain in G0 for long periods of time. Because it can last for an essentially indefinite period of time, resting phase is often considered not to be a proper part of the cell cycle itself. Interphase is when a cell prepares for division, and it can take up approximately 90% of the time of the cell cycle. Two major things happen during interphase: growth and DNA replication. However, interphase is broken into three stages: Gap 1 (G1), synthesis (S), and Gap 2 (G2). During G1 and G2, the cell grows, and during S, DNA is replicated. The fact that S is located between G1 and G2 allows checkpoints. The G1/S checkpoint, also known as the restriction point, is when a cell commits to division. The presence of DNA damage or other external factors can cause a cell to fail this checkpoint and not divide. The G2 checkpoint that takes place before cell division similarly checks for DNA damage after DNA replication, and if damage is detected, serves to "pause" cell division until the damage is repaired. Throughout interphase, chromatin is loosely packaged (euchromatin) to allow transcription and replication. After interphase, the cell undergoes division (mitosis in non-sex/germ cells). Mitosis proceeds through prophase (where the nuclear membrane disappears, chromosomes condense, and the mitotic spindle forms), metaphase (where chromosomes line up along the metaphase plate), anaphase (where chromosome are pulled apart), and telophase/cytokinesis (where the nuclear envelope and nucleolus reappear and the cell divides). Meiosis occurs in sex/germ cells and turns a diploid (2n) parent cell into 4 haploid (n) daughter cells in a two-stage process, in which crossover between homologous chromosomes and the random allocation of maternal/paternal chromosomes to daughter cells work together to create genetic variability.
Cell Cycle and Checkpoints
Fertilization takes place in the Fallopian tube, when a sperm cell encounters a secondary oocyte. The sperm cell passes through the corona radiata, a layer of follicular cells surrounding the oocyte, and the zona pellucida, a layer of glycoproteins between the corona radiata and the oocyte. This triggers the acrosome reaction, in which digestive enzymes are released that allow the nucleus of the sperm cell to enter the egg. The secondary oocyte completes meiosis II, creating a second polar body and a mature ovum. Then, the haploid nuclei of the sperm cell and the ovum merge, creating a diploid one-cell zygote. As the zygote travels to the uterus, it undergoes a series of mitotic cell divisions known as cleavage. Once the zygote has cleaved into a mass of 16 cells by three to four days after fertilization, it is known as the morula. By three to five days after fertilization, the morula develops some degree of internal structure and becomes a blastocyst, with a fluid-filled cavity in the middle known as the blastocoel. The blastocyst implants in the uterine endometrium and further differentiates into the gastrula. The gastrula has three layers: the ectoderm, the mesoderm, and the endoderm. These layers eventually go on to form specific organs and components in the body. The ectoderm primarily gives rise to the nervous system and epidermis (skin), as well as related structures like hair, nails, and sweat glands, and the linings of the mouth, anus, and nostrils. The process through which the nervous system is formed from the ectoderm is known as neurulation. The mesoderm generates many of the structures present within the body, including the musculature, connective tissue (including blood, bone, and cartilage), the gonads, the kidneys, and the adrenal cortex. The endoderm is basically responsible for the interior linings of the body, including the linings of the gastrointestinal system, the pancreas and part of the liver, the urinary bladder and part of the urethra, and the lungs.
Embryology
D is correct. Neural tube defects are defects in the central nervous system. The nervous system is derived from the ectoderm. It can be concluded from the given information that folate is important for neurulation, or the induction of the ectoderm to differentiate into the nervous system. Remember, the ectoderm (shown below) is the outer layer of the gastrula. It gives rise to the nervous system, epidermis, hair, nails, teeth, and sweat glands. A: The maternal nervous system does not communicate directly with the fetal nervous system. In addition, if the fetus already had a nervous system, folate would not help prevent neural tube defects. B: While folate is important for red blood cell development, this information would not explain why its presence helps to prevent neural tube defects. C: The mesoderm gives rise to structures such as bone, cardiac muscle, skeletal muscle, smooth muscle, and tissues involved in the excretory and reproductive systems. Folate is not crucial for this development directly, and the information presented in the question stem does not support this statement.
Folate is a common supplement given to pregnant women to prevent neural tube defects. The most likely reason for this is: A. folate is involved in action potential transmission from mother to fetus. B. folate is directly involved in erythropoiesis. C. folate is critical for proper development of the mesoderm. D. folate plays a role in ectodermal induction.
Gene Expression Promoters are regions of DNA that lie upstream to a given gene and initiate transcription by binding specific transcription factors that contribute to the binding of RNA polymerase. Additionally, expression is upregulated by enhancers, which are DNA sequences that can be located further from the gene of interest, and work by binding transcription factors that twist DNA into a hairpin loop, bringing distant regions into close proximity for transcription to begin. Silencers are the opposite of enhancers in eukaryotic cells; they are regions of DNA to which transcription factors known as repressors bind. Additionally, the methylation of C and A residues can reduce transcription. Methylation is associated with epigenetics, which refers to inheritable phenotypic changes involving mechanisms other than the alteration of the genome itself. Gene expression can also be regulated on the level of nucleosomes (i.e. chromatin and histones). Acetylation promotes transcription by attaching acetyl groups to lysine residues on histones, making them less positively-charged and causing a looser wrapping pattern that allows transcription factors to access the genome more easily. Finally, non-coding RNA plays a role in gene expression. MicroRNA (miRNA) strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while small interfering RNA (siRNA) molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and silence genes by interrupting expression between transcription and translation
Gene Expression
C is correct. Given that ΔG° is negative, and using the equation ΔG° = -RTlnK, ln(K) must be positive for ΔG° to be negative. K > 1. A: This would correspond to a positive ΔG°. B: This would correspond to ΔG° = 0. D: This statement is untrue. ΔG° and K are related according to the equation ΔG° = -RTlnK.
Given that ΔG° is negative for a particular reaction, what can be said about the equilibrium constant K? A. K < 1 B. K = 1 C. K > 1 D. Nothing can be determined about the equilibrium constant K.
D is correct. The modifications were first mentioned in the first paragraph ("acetylation of lysine residues, methylation of lysine and arginine..."), and then a detailed example was given in the last paragraph. These are all modifications to amino acids and/or proteins. Protein modifications must be post-translational modifications. Such modifications can take place in a variety of locations within the cell, such as the interior of the endoplasmic reticulum or the cytoplasm. In contrast, post-transcriptional modifications (those performed on mRNA) occur in the nucleus; these include the addition of the poly(A) tail, the addition of the 5' cap, and splicing. A: Read the passage and answer choices carefully! The modifications mentioned in the final paragraph were made to histones, which are not enzymes (although they are proteins). Choice A is therefore untrue; while post-translational modifications certainly can be made to enzymes, they can also be made to proteins that lack enzymatic function. B: Post-transcriptional modifications do involve catalysis by enzymes. For example, addition of the poly(A) tail is catalyzed by a poly(A) polymerase. C: This statement is false; both the poly(A) tail and the 5' cap require addition to the original pre-mRNA substrate, and splicing involves the removal of introns.
How do the chemical modifications described in the passage differ from eukaryotic post-transcriptional modifications? A. Post-transcriptional modifications are made to mRNA, whereas post-translational modifications only affect enzymes. B. Post-transcriptional modifications are not catalyzed by enzymes. C. Post-transcriptional modifications do not involve the addition or removal of parts of the substrate. D. Post-transcriptional modifications are carried out entirely within the nucleus.
B is correct. During typical breathing, the diaphragm contracts, flattening it and increasing the volume of the intrapleural space. As a result, the lungs expand and the pressure inside them decreases. The resulting pressure differential between the lungs and the atmosphere causes air to rush in, a process termed inspiration. If the lungs are punctured, air will flow freely between the lung and the intrapleural space, and the lung will not expand (since no pressure differential can be maintained). Expansion of the thoracic cavity should still lead to some air flow into the lung, however, although it will also mix freely with the thoracic cavity's air, making it more difficult to absorb oxygen within the damaged lung while also making it harder to maintain a pressure differential with the other lung. Thus, B is definitely true.
If a lung is punctured, what effect will this likely have on breathing? A. The lung will fill in with blood from the surrounding intrapleural pace, leaving less room for oxygen. B. The punctured lung will be unable to inflate, affecting inhalation. C. Breathing will be painful, but otherwise unaffected. D. All air will rush out of the lung and inhalation will be impossible.
A is correct. This question asks us to determine the effects of a change in the GLUT-1 gene. We learn from the passage that GLUT-1 acts mainly in the blood-brain barrier to move glucose into the brain, where it is used more than any other place in the body. The rest of the body will not be directly affected as the brain will be. In fact, De Vivo disease (also known as GLUT-1 deficiency syndrome) is a rare condition caused by inadequate transportation of glucose across the blood-brain barrier, resulting in developmental delays and other neurological problems. B, C: Diabetes is unlikely to result because the body will still be able to take glucose up from the bloodstream in the other tissues of the body. D: There is no reason to think that this would occur unless GLUT-2 transporters functioned in the same tissue as GLUT-1 transporters, which is not the case.
If a person has a mutation in the GLUT-1 gene that results in diminished transport capacity, which of the following is most likely to result? A. Cognitive difficulties B. Type I diabetes mellitus C. Type II diabetes mellitus D. An increase in the production of GLUT-2 transport proteins in the body
B is correct. This question asks us to consider the structure of morphine given in the passage. What needs to change for morphine to enter the brain while maintaining its function? We are told in the passage that lipid-soluble molecules are able to pass through, while hydrophilic molecules are not. Morphine is hydrophilic in part because of its hydroxyl groups. So, replacing them with acetyl groups would allow the drug to become more lipophilic; although the C=O bond in the acetylated molecule is polar, it cannot undergo hydrogen bonding, so the acetylated structure is effectively much more lipophilic than the original structure with -COOH groups. Interestingly, diacetylated morphine is heroin. The effects and pharmacology of heroin are somewhat distinct from those of morphine, although they belong to the same general class of opiates, and heroin does indeed more readily cross the BBB. A: This would only make the molecule bigger and do little for its polarity. Additionally, while glucose can pass through the blood-brain barrier, adding glucose to another molecule would not necessarily allow that molecule to utilize glucose transport proteins. C, D: These would destroy the molecule's structure, making it unlikely to maintain its therapeutic function.
If pharmacologists wished to convert morphine into a form available to brain tissue, which of the following changes could be made to its molecular structure to allow for the best chance to use the drug as a direct brain treatment? A. Attach a glucose molecule to the nitrogen atom B. Replace the alcoholic protons with acetyl groups C. Lyse the molecule to make it smaller D. Remove all double bonds from the molecule
D is correct. DNA sequences that are common among different species, phyla, or even kingdoms are called conserved sequences. Conserved sequences tend to remain that way due to the fact that they code for a vital function that is common among disparate species. A: Not all kingdoms utilize photosynthesis. B: Not all kingdoms engage in cholesterol synthesis. C: Protein modification is something that is quite unique to each phylum or even species, since it is one of the major ways in which species tailor protein function to their needs and create additional protein diversity from the limited set of genes they contain.
If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which of following would be expected to be most similar? A. Photosynthesis B. Cholesterol synthesis C. Protein modification D. DNA synthesis
C is correct. This question requires outside knowledge about glucose metabolism. If cells cannot take up glucose, it will remain in the blood and eventually be excreted in the urine when it builds up to the point that it cannot be reabsorbed by the nephron. In a state of extended hyperglycemia, the body relies on fat metabolism to generate energy, which produces ketone bodies that are also excreted in the urine. Therefore, Roman numerals II and III are correct. I: Proteins in the urine are not the result of hyperglycemia, but rather damage to the glomerulus. Thus, although individuals with diabetes can have protein in their urine, it is not the direct result of elevated glucose levels, but instead a later complication.
In severe diabetic hyperglycemia (high blood sugar), insulin cannot effectively induce the uptake of glucose by cells. Chronic hyperglycemia directly leads to the presence of which of these molecules in the urine? I. Proteins II. Glucose III. Ketone bodies A. I only B. I and II only C. II and III only D. I, II, and III
B is correct. According to the passage, binding of endocannabinoids to CB1 produces "pain-diminishing analgesic" effects. Activation of sensor neurons known as nociceptors by noxious stimuli mediates the perception of pain. A: Chemoreceptors, including those of the taste buds in the tongue, olfactory receptors in the nose, carotid and aortic bodies, and chemoreceptor trigger zone of the medulla, are neurons involved in sensing molecules dissolved in gases or liquids. C: Baroreceptors are a subclass of mechanoreceptors located in the walls of blood vessels, as well as the atrial and ventricular walls of the right side of the heart. They detect stretch, signaling changes in luminal pressure. D: Mechanoreceptors are neurons that respond to mechanical pressure or distortion. Cutaneous mechanoreceptors are responsible for somatosensation, while those embedded in muscles and ligaments are responsible for sensing muscular stretch and load, including in the afferent arm of a number of reflex arcs.
Information in the passage indicates that signaling via CB1 in mice decreases the sensory response initiated by the activation of what receptor type? says that it recieves pain A. Chemoreceptors B. Nociceptors C. Baroreceptors D. Mechanoreceptors
C is correct. Meiosis I results in 2 haploid cells, each with 23 chromosomes consisting of 2 sister chromatids per chromosome. In the male, the sister chromatids are split into 4 gametes during meiosis II. For females, meiosis I results in a secondary oocyte (a gamete) and a polar body. Penetration of the secondary oocyte by a sperm brings on anaphase II. Telophase II produces a zygote and a second polar body. Remember for the MCAT: mitosis results in diploid daughter cells, while meiosis results in haploid cells to produce gametes. A: Neither meiosis nor mitosis results in 46 chromosomes with 2 sister chromatids. In mitosis, the sister chromatids separate to create daughter cells with 46 chromosomes, each chromosome consisting of a single chromatid. B: Haploid cells have 23 chromosomes. D: Diploid cells have 46 chromosomes.
Meiosis I results in: A. 2 diploid cells with 46 chromosomes, each chromosome consisting of 2 sister chromatids. B. 2 haploid cells with 46 chromosomes, each chromosome consisting of 1 chromatid. C. 2 haploid cells with 23 chromosomes, each chromosome consisting of 2 sister chromatids. D. 2 diploid cells with 23 chromosomes, each chromosome consisting of 1 chromatid
B is correct. Competitive inhibitors bind to the active site of the enzyme in question, blocking the substrate from attaching. The question stem states that ethanol is "preferentially processed," or processed by one or more enzymes instead of methanol. Thus, it is highly likely that ethanol binds to the active site where methanol otherwise would, making it a competitive inhibitor.
Methanol poisoning occurs when the body converts a large amount of methanol to harmful chemicals that attack the optic nerves, leading to acute blindness. Ethanol is the antidote for methanol poisoning because ethanol is preferentially processed by the body. Ethanol is most likely to be a(n): A. allosteric inhibitor. B. competitive inhibitor. C. noncompetitive inhibitor. D. irreversible inhibitor.
Mitosis and Meiosis In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis. In contrast, meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell. In prophase I of meiosis, homologous chromosomes (i.e., the maternal and paternal copies of a given chromosome) pair up with each other in a process known as synapsis, forming tetrads. While paired up, homologous chromosomes may exchange genetic information in a process known as crossing over. The crossing-over points are known as chiasmata. This process results in recombinant DNA that is another source of variation in sexual reproduction, in addition to the variability inherent to the process. In metaphase I, homologous pairs, which take the form of tetrads, line up at the metaphase plate. The orientation of the homologous pairs is random in terms of which side of the metaphase plate the maternal or paternal copy of a given chromosome in a homologous pair winds up. In anaphase I, the homologous pairs are separated, and one member of each pair is pulled to each side of the cell. In meiosis II, which operates similarly to mitosis, the sister chromatids are split up into two haploid daughter cells.
Mitosis and Meiosis
B is correct. For the most part, bacteria lack introns, whereas large eukaryotic genes usually contain several introns. A large, eukaryotic gene without introns suggests a bacterial origin. A: The researchers found that not only does the mRNA lack introns (expected for a eukaryote), but the gene itself lacks introns. C: Most bacterial genes do not contain introns. D: As stated in the explanation above, this finding supports the hypothesis that the gene was transferred from Wolbachia to Aedes.
Of what significance to the experimental goals is the finding that AAEL004181 has no introns? A. It is expected, since introns are excised when heterogeneous nuclear RNA is processed into messenger RNA. B. It suggests that AAEL004181 originated from bacteria. C. It is irrelevant, since both bacterial genes and eukaryotic genes contain introns. D. It challenges the hypothesis that AAEL004181 has a bacterial origin.
C is correct. Phenylalanine (shown below) is a benzene-containing amino acid. Children with phenylketonuria are unable to degrade this amino acid to tyrosine, resulting in a high level of phenylalanine in the body, which can cause organ and neuronal damage. A: The side chain of phenylalanine is nonpolar, not polar. B: The side chain of phenylalanine is not acidic. This would only be true of acidic amino acids, like glutamic acid and aspartic acid. D: All amino acids contain hydrogen, so this answer is incorrect. (Even if this choice were referring only to the side chain, it would be describing glycine, not phenylalanine.)
Phenylketonuric children are advised to avoid specific foods containing phenylalanine due to a dysfunction in the degradation of amino acids that contain: A. a polar side chain. B. an acidic side chain. C. a benzene ring. D. a hydrogen group.
The process of going from DNA to RNA—more specifically, messenger RNA (mRNA)—is called transcription. Transcription takes place in the nucleus, and it results in the creation of an mRNA copy of a gene that can then be transported to the cytosol for translation into a protein. The DNA helix must be unzipped for transcription to take place, which means that some of the same machinery used for DNA replication has to be engaged, especially enzymes like helicase and topoisomerase. RNA polymerase is the enzyme responsible for RNA synthesis. In eukaryotes, it binds to a promoter region upstream of the start codon with the assistance of transcription factors, the most important of which is the TATA box. RNA polymerase travels along the template strand in the 3'-5' direction, synthesizing an antiparallel complement in the 5'-3' direction. The template strand is known as the antisense strand, and the opposite strand is known as the sense strand, because it corresponds to the codons on the mRNA that is eventually exported to the cytosol for translation. The immediate product of transcription in eukaryotes is not mRNA, but heterogeneous nuclear RNA (hnRNA). hnRNA must undergo a set of post-transcriptional modifications to become mRNA. Examples commonly tested on the MCAT include the 3' poly-A tail, the 5' cap, and splicing. The 3' poly-A tail is a string of approximately 250 adenine (A) nucleotides added to the 3' end of an hnRNA transcript to protect the eventual mRNA transcript against rapid degradation in the cytosol. The 5' cap refers to a 7-methylguanylate triphosphate cap placed on the 5' end of an hnRNA transcript. Similarly to the 3' poly-A tail, it helps prevent the transcript from being degraded too quickly in the cytosol, but it also prepares the RNA complex for export from the nucleus. In splicing, noncoding sequences (introns) are removed and coding sequences (exons) are ligated together. (Remember that exons are expressed). Each gene normally has multiple distinct exons that can be ligated in different combinations; that is, if a gene had a set of four exons named A, B, C, and D, possible alternate splicing combinations could include ABCD, ABC, ACD, ABD, BCD, and so on. This dramatically increases the amount of different, but related proteins that can be expressed from a single gene. Splicing explains why there are over 200,000 proteins in the human body, but only approximately 20,000 genes. Splicing is carried out by the spliceosome, a combination of small nuclear RNAs (snRNAs) and protein complexes.
RNA Transcription
A is correct. Lysine is a basic amino acid and typically has a positively-charged side chain at physiological pH. When lysine is acetylated, this charge becomes neutral, as shown in the figure below. Since DNA is negatively charged due to its phosphate backbone, the charge on lysine allows for tight histone-DNA interactions thanks to electrostatic attraction between the charged atoms on each molecule. Acetylation of lysine makes the residue neutral, lessening these interactions and promoting a looser structure. Loose chromatin structure is typically associated with euchromatin, the less dense, transcriptionally active chromatin structure that appears light under a microscope. In contrast, histone deacetylation will restore the positive charge to the residue, allowing the electrostatic attractions to return. Therefore, deacetylation of lysine residues on histones should lead to a denser chromatin structure and lowered transcription/gene expression (choice A). B, D: Both of these statements are the reverse of the typical effect of histone acetylation. C: Gene expression and histone modifications impact transcriptional activity, not replication. A gene should be replicated during each S phase regardless of the post-translational modifications on its histones.
Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge. Combined with the information in the passage, which of the following conclusions will the researchers most likely reach? A. Histone deacetylation generally decreases gene expression. B. Histone acetylation tends to promote a more condensed chromatin structure. C. Histone deacetylation reduces the extent to which the gene in question will be replicated. D. Histone acetylation favors the formation of heterochromatin rather than euchromatin.
C is correct. If the scientists wanted to prevent cellular replication, they would need to halt cell division (mitosis). Interphase is the stage of the cell life cycle (shown below) that occurs between rounds of division. A, B, D: Prophase, metaphase, and anaphase are all phases of mitosis. If cells were arrested at any of these stages, they would already have begun to replicate
Scientists who wished to study the metabolic function of cells with balanced translocations while preventing cell replication would be best served by arresting the cells during which phase of the cell cycle? A. Anaphase B. Metaphase C. Interphase D. Prophase
Spontaneity A spontaneous reaction is one that can proceed without the help of some additional, external force. Importantly, spontaneity does not relate to the rate of the reaction, or the speed at which it progresses—spontaneous reactions can be very slow! Rate is a kinetic parameter that is determined largely by the activation energy. In contrast, spontaneity is a thermodynamic parameter and relates to ∆G, or the change in Gibbs free energy. The most basic definition of spontaneity is ∆G < 0, meaning that the reaction has a negative change in Gibbs free energy. Spontaneity is associated with exothermic (∆H < 0) reactions and those that increase entropy (∆S > 0) through the equation ∆G = ∆H - T∆S. Note that a reaction can still be spontaneous even if it fulfills only one of these requirements. For example, a spontaneous reaction may be endothermic (∆H > 0) as long as this is balanced by a sufficiently positive ∆S. Spontaneity can also be defined in terms of other chemical concepts. By definition, a spontaneous reaction will lead to more products being present than reactants. This means that the equilibrium constant (Keq), which is broadly defined as [products]/[reactants], will be greater than 1 for a spontaneous reaction. This relationship is encoded in the equation ∆G°rxn = −RTlnKeq. Additionally, in the context of electrochemical cells, spontaneous reactions are associated with positive cell potentials (E° > 0). On Test Day, it is important to automatically recognize that spontaneity is equivalent to ∆G < 0, Keq > 1, and E° > 0.
Spontaneity
C is correct. Rapidly dividing cells undergo mitosis under the influence of specific signaling molecules. These molecules are expressed when their genes are transcribed, then are translated into proteins. In order to gain the best understanding of how a signaling protein's levels are regulated, both the protein and mRNA levels would need to be studied. Western blotting gives us information about the amount of protein expressed in a cell, while RT-PCR gives us information about the amount of RNA expressed. A, B, D: Southern blots are used to probe DNA for specific sequences; however, changes of gene expression at a cellular level are due to changes in the transcription and translation rates for the genes, not the number of genes themselves.
The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins? A. RT-PCR and Southern blot B. Southern blot and northern blot C. Western blot and RT-PCR D. Western blot and Southern blot
B is correct. The cell membrane is composed of several different components, each responsible for different functions. Membrane transport is most likely to be affected if the disruption occurs in components that span the entire membrane. Transmembrane proteins (many of which are glycoproteins) are the only component listed that pass all the way through the cell membrane and facilitate membrane transport. A: Cholesterol does not facilitate membrane traffic, nor does it extend across the entire membrane. C, D: Phospholipids and glycolipids are located on the surface of cell membranes and typically do not extend though the entire bilayer. Glycolipids act to provide energy and also serve as markers for cellular recognition. Phospholipids are a structural component of the membrane and are not involved in traffic/transport.
The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders? A. Cholesterol B. Glycoproteins C. Glycolipids D. Phospholipids
D is correct. At a pH of 8, all of the molecules as free amino acids will have at least one positive and one negative charge, as the carboxylic terminus will have a -1 charge and the amino terminus will have a +1 charge. According to the given table, arginine and glutamic acid will also have charged side chains.
The table below gives pKa data for selected amino acids. Which of the amino acids listed will exist predominantly as a molecule with both positive and negative charges at pH of 8.0? I. Arginine II. Tyrosine III. Glutamic acid IV. Glycine A. IV only B. I and III only C. II and IV only D. I, II, III, and IV
D is correct. Be sure to read the question stem carefully! Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH-). Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would not be affected, and the curve would fall along line 1. A: This would represent a rate higher than that of the uninhibited reaction, which does not make sense here. B, C: These choices represent inhibited reactions, with rates lower than that of the uninhibited reaction. In reality, however, the two inhibitors would neutralize each other before being added to the substrate-containing mixture, and no inhibition will occur.
Two known competitive inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Based on Figure 1, where would the resulting enzyme kinetics curve for this experiment fall? A. Above line 1 B. Below line 1 but above line 3 C. Below line 3 D. The same as line 1
C is correct. Uracil is found in any structure made of RNA. Both tRNA and ribosomes are made of RNA. Single-stranded DNA has thymine rather than uracil.
Uracil is usually found in: I. tRNA. II. ribosomes. III. single-stranded DNA. A. I only B. II only C. I and II only D. I, II, and III
B is correct. Isopropyl alcohol is a three-carbon chain with a hydroxyl group on the middle carbon: CH3 - CHOH - CH3 Alternatively to writing out the formula (as shown above), you could draw out the chemical structure of isopropyl alcohol. (Note that the central atom in the diagram below is bound to an "invisible" hydrogen atom.) Either method shows that the formula for isopropyl alcohol should be C3H7OH, since one H atom is replaced by one -OH group. The best approach for solving a problem like this one is to draw the relevant diagram. A: Here, there are too many H atoms for the carbons present. C: Isopropyl alcohol contains only one -OH group, not two. D: This formula does not contain enough H atoms.
What is the correct molecular formula for isopropyl alcohol? A. C3H8OH B. C3H7OH C. C3H6(OH)2 D. C3H6OH
D is correct. Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein. A, B: These controls occur before transcription. In contrast, siRNA prevents mRNA from being translated. C: Repressors are defined as protein molecules that bind with DNA or RNA to prevent eventual translation of a protein. Therefore, siRNA is not technically a repressor.
What type of control does siRNA exert on G6PD expression? A. Transcriptional control B. Promotion C. Repression D. Post-transcriptional control
B is correct. The answer is given directly in paragraph 2, which mentions "...two hemichannels (connexons), one embedded in each cell's plasma membrane and anchored to the cytoskeleton of the cells." This means that the cytoskeleton of one cell is connected to the cytoskeleton of another. A schematic of a gap junction is shown below. A, D: Gap junctions don't connect to the nucleus. C: The cytoskeleton of one cell physically connects to the connexons, which then physically link to the cytoskeleton of the next cell. Gap junctions don't bring the cytoskeleton of one cell into contact with the plasma membrane of the neighboring cell.
Which components of cells are physically connected by a gap junction? A. The cytosol of one to the nucleus of the other B. The cytoskeleton of one to the cytoskeleton of the other C. The cytoskeleton of one to the plasma membrane of the other D. The plasma membrane of one to the nucleus of the other
B is correct. This answer relates to the differences between these modes of transfer between cells. At gap junctions, the cells approach within about 2-4 nm of each other, a much shorter distance than the 20-to-40-nm intracellular space that separates cells at a chemical synapse. In chemical synapses, the chemicals must diffuse over these larger distances, which takes more time. A chemical synapse is shown below; note that it is what we typically think of when we think of a synapse. Indeed, chemical synapses are more common than electrical synapses. A, C: According to the information in the passage and our knowledge of skeletal muscle action potentials and neuronal AP conduction, these statements are false. D: This choice is true, but a lack of signal modification does not explain the difference in speed between the two synapse types. It also does not relate to the large disparity in distance the signal needs to travel.
Which of the following best explains why chemical synapses are slower than electrical synapses? A. Chemical synapses are not initiated by an electrical signal. B. Chemical synapses require the movement of agents through intercellular space. C. Electrical synapses require the movement of chemicals through a small opening. D. Electrical synapses lack gain; the signal in the postsynaptic neuron is the same or smaller than that in the originating neuron.
D is correct. A nucleoside is composed of a nitrogenous base and a five-carbon sugar (ribose or deoxyribose). Adenosine, shown below, is such a molecule. For adenosine, the nitrogenous base is adenine, while the five-carbon sugar is ribose (note the presence of both a 2' and a 3' hydroxyl group). A, B: IMP and GTP are both nucleotides. Nucleotides are composed of a nitrogenous base, a five-carbon sugar, and at least one phosphate group. C: Hypoxanthine is a nitrogenous base.
Which of the following biomolecules is an example of a nucleoside? A. IMP B. GTP C. Hypoxanthine D. Adenosine
C is correct. From outside knowledge, we know that insulin secretion increases after a meal to help the cells take up glucose from the bloodstream. The passage states that the PPP is a parallel (alternative) path to glycolysis, so it makes sense that it would also be activated when blood sugar levels rise after a meal. A, B: Insulin secretion increases after eating. D: G6PD activity would not decrease after carbohydrate consumption
Which of the following changes occur immediately following the consumption of a carbohydrate-rich meal? A. Insulin secretion decreases; G6PD activity increases. B. Insulin secretion decreases; G6PD activity decreases. C. Insulin secretion increases; G6PD activity increases. D. Insulin secretion increases; G6PD activity decreases.
B is correct. This question is asking us to recall some facts about mitosis and meiosis. Remember that mitosis separates sister chromatids to create two diploid daughter cells. Meiosis I separates homologous chromosomes to create haploid daughter cells, each of which divides again, separating sister chromatids in Meiosis II to create two haploid cells. A: Mitosis results in two diploid daughter cells. C: During meiosis I, sister chromatids are not separated. D: Meiosis I results in two haploid daughter cells.
Which of the following is an accurate statement regarding mitosis and meiosis? A. Mitosis results in two haploid daughter cells. B. During mitosis and meiosis II, sister chromatids are separated. C. During meiosis I and meiosis II, sister chromatids are separated. D. Meiosis I results in two diploid daughter cells.
D is correct. In each case, you should draw the Lewis dot structure. In methanol (CH3OH), four atoms form single bonds with the carbon atom, which is sp3 hybridized. The oxygen atom has two single bonds, one to carbon and one to hydrogen, as well as two lone pairs of electrons; it therefore has sp3 hybridization as well. Neither atom has sp2 hybridization. The structure of methanol is shown below. A: In the case of carbon dioxide (below), the carbon is sp hybridized. However, since there are two lone pairs on the oxygen atoms and a double bond between the carbon and each oxygen atom, the oxygen atoms are sp2 hybridized. B: In the case of carbonate (CO32-) the carbon atom is sp2 hybridized, as shown below. C: In the case of formaldehyde (CH2=O) both the carbon and the oxygen atoms are sp2 hybridized.
Which of the following molecules does NOT have an atom with sp2-type hybridization? A. Carbon dioxide B. Carbonate C. Formaldehyde D. Methanol
C is correct. The correct answer will be a molecule or substance that is essential to brain function. Amino acids are necessary for the production of proteins, which are essential for the function of any cell. A: In general, antibodies are too large to cross the BBB. B: While select monosaccharides (e.g. glucose, fructose) can cross the BBB, large polysaccharides like starch are unable to cross. D: Urea is a waste product that is unlikely to have specialized proteins for entry into the brain.
Which of the following molecules is/are most likely to have selective proteins in the BBB to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea
B is correct. This question requires us to remember the workings of mitosis. In mitosis, the spindle fibers that move chromosomes are made up of microtubules. They are attached to chromosomes in metaphase and pull them apart in anaphase. Therefore, the first phase to be affected by lack of microtubules will be metaphase. Spindle fibers, along with associated structures, are shown below. A, C: These are phases of meiosis, not mitosis. D: Cytokinesis is the splitting of the cell into two. It is not a phase of mitosis, and also occurs after both metaphase and anaphase.
Which phase of mitosis is likely to be first interrupted if a cell has no microtubules? A. Prophase II B. Metaphase C. Telophase I D. Cytokinesis
C is correct. The best way to approach this question is by process of elimination. It can be helpful to first eliminate answers that are untrue, and then to eliminate true choices that do not answer the question. Reasoning for how to eliminate each option is shown below. Once we eliminate options A, B, and D, we see that choice C is both accurate and relevant to the question stem: this 15 mL/kg of air is the residual volume, which is the minimum volume the lungs can have, and which is attained at maximum intrapleural pressure. A: Remember, work = PΔV, so PV would not be constant in a case where work is being done on or by a mass of air. Additionally, you may be thinking of the ideal gas law, where for a certain number of moles of an ideal gas at constant temperature, PV is constant. During breathing, however, the molar amount of air in the lungs changes over time. B: This is tempting, but read each answer closely. We must know for test day that contraction of the diaphragm flattens the muscle and increases thoracic volume. This facilitates inspiration, not expiration (as would be expected to explain minimum lung volumes). Forced exhalation beyond the norm uses not the diaphragm muscles (which simply relax during expiration), but the intercostal muscles. D: This is true because if air was forcibly removed from the lungs, below a certain point, the lungs would collapse under intrapleural pressure. However, this will not occur spontaneously from exhaling with too much force. That leaves only C, the sole working statement. Pressure and volume are inversely related, so under maximum pressure, the lungs will reach a minimum volume. This volume of air remaining in the lungs is termed the residual volume.
Which statement best explains why the lung volume in Figure 1 never drops below 15 ml/kg? A. Since PV is constant, pressures below that value will cause ruptures in the lung tissue. B. At this volume, maximum contraction of the diaphragm has forced out as much air as possible from the lungs. C. The residual volume is such that the lungs are at their minimum volume under maximum intrapleural pressure. D. Below this volume, spontaneous lung collapse will occur, and the body has evolved safeguards.
Two distinct forms of chromatin exist: euchromatin and heterochromatin. Euchromatin is a loose configuration that is difficult to see under light microscopy and allows DNA to be readily transcribed. Throughout interphase (i.e., most of the cell cycle), DNA generally exists as euchromatin. Heterochromatin is the tightly coiled, dense form of chromatin that is visible during cell division and is present to a lesser extent even during interphase.
euchromatin and heterochromatin
Histones are proteins that act as spools for DNA to wind around. They are composed of various subunits known as H1, H2A, H2B, H3, and H4. The core of a histone contains two dimers of H2A and H2B and a tetramer of H3 and H4, while H1 serves as a linking unit. Approximately 200 base pairs of DNA can be wound around a histone, and the complex formed by DNA and a histone is known as a nucleosome. The phrase "beads on a string" is often associated with the appearance of nucleosomes under electron microscopy, and chromatin refers to the structure formed by many nucleosomes. On a biochemical level, charge plays a major role in the interactions between histones and DNA. Histones are highly alkaline and are positively charged at physiological pH, which facilitates their interaction with the highly negatively charged phosphate groups on the backbone of DNA. Modifications like acetylation of histones reduce that positive charge, making histones interact with DNA less closely, which in turn facilitates transcriptional activity.
histones
Aneuploidy results from having too many or too few copies of a given chromosome. This results from nondisjunction in anaphase during cell division. Having only one copy of a chromosome is known as monosomy, and having three copies is known as trisomy. Aneuploidy is commonly discussed as occurring in meiosis, and indeed, this is the only way for aneuploidy to be inheritable. For this reason, nondisjunction during meiosis is the cause of aneuploidies such as Down syndrome (trisomy 21) or Turner syndrome (monosomy X). However, nondisjunction during mitosis can also occur, and this is extremely common in cancer cells.
when and where does Aneuploidy occur