Incorrect Biology Problems

In a given population, what is the frequency of carriers of Patau syndrome? A) 1/100 B) 198/10,000 C) 9,801/10,000 D) None of the above

D. The Hardy-Weinberg equation can only be used to describe the frequencies of autosomal recessive or dominant traits or conditions, not changes in chromosome number. Since trisomies are not autosomal recessive conditions, the Hardy-Weinberg equation cannot be used, and one cannot predict the occurrence of the carriers in a population (choice D is correct and choices A, B, and C are eliminated). Also, because trisomies are not based on dominant or recessive expressions, there is no "carrier" state.

What component of a cell membrane is most likely to be affected in a membrane trafficking disease? Please choose from one of the following options. Peripheral proteins Glycoproteins Integral proteins Cholesterol

The cell membrane is composed of several different components, each responsible for different functions. Hint #2 Remember that trafficking or membrane transport is likely mediated by components that go through the entire the membrane. Hint #3 Integral proteins are the only component that affect membrane transport and they pass all the way through the cell membrane.

Overdosing on acetylsalicylic acid can cause a metabolic acidosis as well as hyperventilation. Which of the following is true? A) Blood pH will likely be 7.2. B) Little compensation is needed to correct the acid-base disturbance. C) Blood pH will likely be 7.6. D) The kidneys will need to secrete more HCO3- to compensate for the acid-base disturbance.

B. A metabolic acidosis is caused by a decrease in HCO3- concentration, resulting in a comparatively higher than normal H+ concentration. Hyperventilation causes a decrease in CO2 concentration. Remembering that CO2 is acid (see Equation 1 and the formation of H+), hyperventilation creates a respiratory alkalosis. The combination of both respiratory alkalosis (which results in a higher blood pH) and metabolic acidosis (which results in a lower blood pH) would result in a near normal blood pH. As a consequence, there will be little compensation required to maintain a normal level pH in the blood (choice B is correct). A pH of 7.2 indicates acidemia, typically caused by an acidosis, or potentially two acidoses (respiratory and metabolic, choice A is incorrect). A pH of 7.6 represents an alkalemia, caused by one or two alkaloses (choice C is incorrect). If the kidneys secreted HCO3-, then the blood pH would become lower since the body is ridding itself of HCO3-, a base. This compensation would only be necessary if there was an alkalemia to correct (choice D is incorrect).

How many possible D isomers of D-ketohexose exist? Please choose from one of the following options. A. 2 B. 4 C. 6 D. 8

First, consider how many chiral centers a ketohexose contains. A ketohexose contains 3 chiral centers, so there are 2^3 = 8 total isomers. The D-isomers account for half of the total isomers, so there are 8/2 = 4 isomers of D-ketohexose.

The conformational change of a regulatory protein after the binding of a repressor most likely represents an alteration of the protein's: A) amino acid composition. B) primary structure. C) secondary structure. D) tertiary structure.

D. A conformational change in a protein is a change in the larger scale folding of a protein, with changes in the relative positions in space of amino acids located far from each other in the linear polypeptide chain. The tertiary structure of a protein involves large-scale structure within a polypeptide chain that is stabilized by interactions between amino acids that can be distant from each other in the linear sequence; thus, this is the most likely level of structure altered during a conformational change (choice D is correct). The amino acid composition and primary structure (the linear sequence of amino acid residues in the polypeptide chain) cannot be changed without breaking covalent peptide bonds, which does not occur in a conformational change (choices A and B are wrong). The secondary structure of a protein involves folding that is stabilized by the nearest neighbors in the polypeptide chain, including structures such as α helix and β sheet. Conformational changes can alter secondary structures somewhat, but mostly alter tertiary structure (choice D is a better answer than choice C).

How many times do opioid receptors cross the plasma membrane? Once Three times Seven times Nine times

Opioid receptors are a type of g-protein coupled receptors. Hint #2 G-protein coupled receptors are also known as "seven transmembrane receptors." Hint #3 The distinguishing structural feature of g-protein coupled receptors is that it is made up of 7 alpha helices that cross our plasma membrane seven times.

What is the C-5 epimer of L-Talose? Please choose from one of the following options. A. D-Mannose B. D-Galactose C. D-Allose D. L-Gulose

Remember, an epimer is a sugar diastereomer that varies at one chiral center. In this case, the specific chiral center is Carbon5 Hint #2 L-Talose is the enantiomer of D-Talose. That means it is a non-superimposable mirror image and will differ at every chiral center. Hint #3 D-Allose. If you compare the enantiomer of D-Talose (L-Talose) with D-Allose, the only carbon with different chirality is Carbon5, making D-Allose the C-5 epimer of L-Talose.

If the researcher in the passage is correct about increased albumin concentration in the interstitial fluid of the brain following focal ischemia, which of the following interactions would most likely be affected? Please choose from one of the following options. Astrocytes and blood vessels Schwann cells and axons Microglia and lymphocytes Ependyma and cerebrospinal fluid

Schwann cells form the myelin sheath around axons of neurons in the peripheral nervous system. Hint #2 Microglia present antigens to lymphocytes. Hint #3 Astrocyte end feet surround blood vessels of the central nervous system to contribute to the blood brain barrier.

What type of protein would NPC1L1 protein likely be? Please choose from one of the following options. Integral protein Transcription factor Peripheral protein Scaffold protein

The cell membrane is composed of several different components, each of which is responsible for different functions. Hint #2 Trafficking or membrane transport is likely mediated by components that go through the entire the membrane. Hint #3 Integral proteins are the only component that affect membrane transport/trafficking and they are a membrane component that pass all the way through the cell membrane.

Which cell type normally maintains homeostasis of the interstitial fluid in the brain, including the concentration of potassium ions? Please choose from one of the following options. Neurons Ependyma Microglia Astrocytes

The ependyma is a leaky barrier composed of ependymal cells between the cerebrospinal fluid and the interstitial fluid of the central nervous system. Hint #2 Neurons process and transmit information in the nervous system. Hint #3 Astrocytes maintain homeostasis of the interstitial fluid of the central nervous system.

"Error-prone reverse transcriptase"

The error-prone reverse transcriptase means that it has a very high mutation rate. This makes it difficult for the immune system to fight every version of it.

Nevirapine is an example of a non-nucleoside reverse transcriptase inhibitor. Which of the following is true of nevirapine? A) It is a competitive antagonist of reverse transcriptase. B) It is a noncompetitive antagonist of reverse transcriptase. C) It is an uncompetitive antagonist of reverse transcriptase. D) Its effect on reverse transcriptase can be reversed with increasing concentrations of adenine, thymine, cytosine, and guanine.

B. Nevirapine is a non-nucleoside antagonist. As the passage states, the non-nucleosides are allosteric inhibitors. As a result, they must be noncompetitive inhibitors of reverse transcriptase (choice B is correct and choices A and C are eliminated). Noncompetitive inhibitors cannot be reversed by increasing the concentration of substrate; that would work only for a competitive inhibitor (choice D is eliminated).

If there were 44 Vh gene segments, 27 Dh gene segments, and 6 Jh gene segments, how many potential sequences (x) can the antibody heavy chain have? A. 44 * 27 * 6 B. x = 1 C. x < 44 * 27 * 6 D. x > 44 * 27 * 6

Because there are 44 Vh gene segments, 27 Dh gene segments, and 6 Jh gene segments, there are 44∗27∗6 different combinations of gene segments that can create an antibody heavy chain. However, additional mechanisms besides V-D-J recombination can contribute to the sequence of the heavy chain. Think about the effects of N-addition, P-addition, and junctional flexibility. Because additional changes can be made to the final antibody heavy chain sequence by N-addition, P-addition, and junctional flexibility, the total potential number of antibody heavy chain sequences is much GREATER than 44∗27∗6.

What is the FRC for an individual with the following lung volumes: total lung capacity (TLC) = 5700 mL, vital capacity (VC) = 4700 mL, inspiratory reserve volume (IRV) = 3000mL and tidal volume (VT) = 500 mL? A) 3500 mL B) 1000 mL C) 2200 mL D) 1200 mL

C. The passage states that FRC = ERV + RV. TLC = VC + RV rightwards arrow RV= TLC - VC = 5700 mL - 4700 mL = 1000 mL VC = VT + IRV + ERV rightwards arrow ERV = VC - VT - IRV = 4700 mL - 500 mL - 3000 mL = 1200 mL FRC = ERV + RV = 1200 mL + 1000 ml = 2200 mL Thus choice C is correct and choices A, B and D are eliminated.

NTKA functions in cells derived from which developmental structure? Please choose from one of the following options. A. Archenteron B. Ectoderm C. Mesoderm D. Blastopore

This type of question is an MCAT favorite! The archenteron forms the primitive gut in the human embryo. The blastopore gives rise to the anus. The mesoderm gives rise to the cardiovascular system, the musculoskeletal system, the lymphatic system, and other structures. The ectoderm gives rise to the nervous system, skin, the lens of the eye, and other structures. Since NTKA is in neurons, this ectoderm is the correct answer.

Given the enzymatic mechanism of NTKA deficiency, what are the most likely genotypes of patients 1 & 2? A. Heterozygous and homozygous dominant B. Homozygous dominant and heterozygous C. Homozygous recessive and homozygous dominant D. Heterozygous and homozygous recessive

* insert answer after * took a screenshot of it

How did the stem cell transplant affect the immune system of the Berlin Patient? A. The transplant allowed the Berlin Patient to manufacture white blood cells, which had been halted by the HIV B. The transplant caused the Berlin Patient to begin to make HIV resistant white blood cells C. The transplant replaced the HIV target cells with the resistant white blood cells from the donor D. The transplant gave the Berlin Patient enough HIV resistant white blood cells to fight both the cancer and the virus

A stem cell transplant injects new progenitor cells (often bone marrow) into the patient. HIV destroys white blood cells, but it does not destroy the host's ability to produce more. A stem cell transplant can change a host's genotype - in the case of the Berlin Patient, after his SCT he began to produce white blood cells resistant to HIV.

A student uses a Geiger counter to measure the radioactivity of a substance with a half-life of 10 days. If he originally recorded a rate of 800 clicks per minute, approximately what click rate should he expect from the same sample 60 days later? A) 12 per minute B) 25 per minute C) 80 per minute D) 130 per minute

A. A time period of 60 days corresponds to 60/10 = 6 half-lives. In this time, the activity rate of the radioactive sample would decrease by a factor of 26 = 64, implying a click rate of 800/64 = 12 per minute.

Protostomes undergo spiral and determinant morula cleavage, where the two first divisions of the zygote are equal and the others are unequal. This results in some small and some large cells early on in morula formation, and means cell fate is decided early in the process. Deuterostomes develop differently. They perform radial cleavage, which results in totipotent cells that are all the same size. Which of the following is supported by this information? A) Humans are deuterostomes. B) The deuterostome morula is more differentiated than the protostome morula. C) The protostome morula cells are more similar to stem cells than the deuterostome morula cells D) The protostome morula has undergone less cell specialization than the deuterostome morula.

A. Based on information in the question, the protostome morula is more differentiated (or specialized) than the deuterostome morula (choices B and D can be eliminated). This means the deuterostome morula cells are more stem cell like (choice C can be eliminated). Cells in the human morula are all the same and are totipotent (choice A is correct).

If the researchers had used a luminescent probe to visualize the location of FAS itself in Experiment 3, where would it predominantly be found? A) Cytosol B) Nucleus C) Mitochondria D) Smooth ER

A. Fatty acid synthesis takes place in the cytosol and requires the presence of FAS (choice A is correct). No part of the fat metabolism takes place in the nucleus of the cell (eliminate choice B). The mitochondrial matrix is the location for beta-oxidation (fatty acid break down; eliminate choice C). Elongation and desaturation of fatty acids takes place inside the smooth ER using separate enzymes. FAS is not used during extensions of fatty acids beyond carbon-16 nor is it involved in the introduction of any double bonds (eliminate choice D).

Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? A) Glycogen phosphorylase B) Glucose-6-phosphatase C) Phosphoglucomutase D) Pyrophosphatase

A. Glycogen phosphorylase catalyzes the rate limiting step of glycogenolysis—the release of glucose-1-phosphate from glycogen via its action on terminal α(1→4) linkages. This makes choice A true and the correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. Choice B is false and an incorrect answer. Phosphoglucomutase is involved in both glycogenolysis and glycogenesis by interconverting glucose 1-phosphate and glucose 6-phosphate, thus choice C is false and the incorrect answer. Pyrophosphatases are members of the acid hydrolase family of enzymes that cleave diphosphate bonds. Choice D is false and an incorrect answer.

B lymphocytes are involved in a process called: A) humoral immunity. B) cell-mediated immunity. C) passive immunity. D) the complement system

A. Humoral immunity refers to immunity through fluid; in this case, the blood. This is where B cells have their effect: They release antibodies into the bloodstream (choice A is true). Cell-mediated immunity is the process that cytotoxic T cells are involved in (choice B is wrong). Passive immunity refers to an injection of antibodies which can act as a temporary supply and help fight off a disease until a person's own immune system can start producing antibodies (choice C is wrong). The complement system is a system of proteins and enzymes which, when activated, can cause cell lysis (choice D is wrong).

In another experiment, female White Leghorn chicks are administered 1.0 mg of stilbestrol for one week in the presence or absence of clomiphene citrate, an estrogen antagonist. Which of the following results would NOT contradict the findings of the first experiment? A) Clomiphene decreased the average oviduct weight from maximum levels. B) Stilbestrol decreased the average oviduct weight. C) Clomiphene has no effect on stilbestrol-induced weight gain. D) Before stilbestrol administration, the test chicks showed maximal levels of weight gain.

A. If stilbestrol is an estrogen receptor agonist (i.e., has effects like those of estrogen) and clomiphene citrate is an estrogen antagonist, then clomiphene citrate should reverse the increase in oviduct weight that is induced by stilbestrol (A is correct, and C is wrong). Statements B and D contradict the passage since Figure 1 shows that stilbestrol increases oviduct weight.

Contraction of cardiac muscle begins just after depolarization and lasts about as long as the action potential. During a contraction: A) Ca2+ enters the cell through voltage-gated ion channels. B) sarcomere length increases. C) polymerized actin becomes depolymerized. D) myosin binds irreversibly to actin.

A. Part of the cardiac action potential is the opening of slow voltage-gated calcium channels to create the plateau in depolarization (Phase 2) so choice A is correct. When the voltage-gated calcium channels open, calcium enters the cell and can play a role in the contraction of the cardiac muscle. Sarcomere length decreases during each contraction, and does not increase (choice B is wrong). Actin filaments in contractile tissue are not dynamic as they are in other processes and do not spontaneously depolymerize and repolymerize (choice C is wrong). Myosin binds actin reversibly, not irreversibly (choice D is wrong).

Cells from the pigmented region of a frog gastrula are transplanted to a light gastrula region. The light region surrounding the graft is induced to form a second incomplete tadpole. Cells transplanted from a light region to the same site do not induce formation of a second tadpole. Based on the results of this experiment, the developmental fate of the region surrounding the graft most likely depends on: A) the region from which the graft cells were taken. B) the region to which the graft cells are transplanted. C) the expression of pigment genes. D) the migration of the graft cells throughout the transplant region.

A. Pigmented cells induce development of an incomplete tadpole, while cells from the light region do not when transplanted to the same area. The region from which the cells are taken appears more important for developmental fate than the area they are transplanted to (choice A is correct, and choice B is wrong). Pigmented cells appear to be the key factor, but pigmentation itself is likely to be coincidental and not the cause of the developmental fate of the region (choice A is a better answer than choice C). There is no evidence presented that migration of cells is important (choice D is wrong).

Which of the following statements about pyruvate kinase are true? A) It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. B) It has quaternary protein structure, similar to hemoglobin and myoglobin. C) It is involved in gluconeogenesis, the conversation of pyruvate to glucose during high energy states. D) It has several isoforms due to alternative splicing via the spliceosome and ribosome.

A. Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP (choice A is correct). Paragraph 3 discusses how PKM2 can exist as a dimer or tetramer, each of which would contain more than one peptide chain and would therefore have quaternary protein structure. Hemoglobin also has quaternary structure (since it contains four peptide chains) but myoglobin does not (eliminate choice B). Since PKM2 is one isoform of pyruvate kinase, there are presumably others. Isoforms are different forms of the same protein and can be due to gene duplication or alternative splicing. However, the ribosome has no function in splicing (eliminate choice D).

Retinal isomerase is used to alter the conformation of retinal. Retinal isomerase is most likely manufactured by: A) dehydration synthesis. B) hydrolytic synthesis. C) oxidation. D) glycosidic anabolism

A. Retinal isomerase, an enzyme, is a protein, and proteins are synthesized by the condensation of amino acids. Condensation involves dehydration.

Aldosterone's mechanism of action on its target cell is most similar to that of which of the following hormones? A) Testosterone B) Glucagon C) Insulin D) ACTH

A. Two general classes of hormones are those that are small hydrophobic molecules like steroid hormones and those that are peptides. The steroid hormones, which include aldosterone and testosterone, diffuse through the plasma membrane to bind to a receptor which enters the nucleus to regulate transcription of a specific set of genes (A is correct). Peptide hormones, such as glucagon, insulin, and ACTH, cannot diffuse into the cell since they are large and hydrophilic, so they bind to cell-surface receptors to transduce a signal into cells (B, C, and D are wrong).

Which of the following describes how the brain uses parallel processing to process a visual stimulus? A) A different area of the brain is activated by each type of image (images, letters, etc.). B) Visual areas V1-V5 are utilized to analyze different aspects of an image simultaneously. C) Images are simultaneously processed using both top-down and bottom-up processing. D) Visual areas V1-V5 are utilized to analyze each aspect of an image in succession.

B. Parallel processing describes the simultaneously processing of different aspects of a stimulus (choice B is correct), instead of analyzing the pieces of a stimulus step-by-step (choice D can be eliminated). Choice A is the definition of feature detection theory, and can be eliminated. Top-down and bottom-up processing are elements of gestalt psychology; the simultaneous use of both top-down and bottom-up processing is characteristic of most sensory processing, but is not the definition of parallel processing (choice C can be eliminated).

Which B-cell process(es) would one expect to occur in a germinal center of a lymph node? I. point affinity maturation II. point B-cell maturation III. point somatic hypermutation IV. point productive rearrangement of Vh, Dh, & Jh. A. I only B. I, II, & IV only C. I & III only D. II & IV only

B-cell maturation occurs in the bone marrow. Hint #2 Somatic hypermutation occurs primarily after a B-cell recognizes antigen and begins dividing. Hint #3 Affinity maturation occurs after initial antigen recognition and selects for B-cells with high binding affinity for antigen. Hint #4 B-cells migrate to germinal centers after activation, where they frequently contact antigen particles that drain into lymph nodes from surrounding tissue. Hint #5 Activated B-cells undergo somatic hypermutation, which provides small differences between B-cells of the same lineage. Those that bind antigen most tightly are selected for in the process of affinity maturation. Hint #6 Since somatic hypermutation and affinity maturation occur after B-cell activation, they are likely to occur in a germinal center of a lymph node.

Diabetes insipidus is a condition in which an individual fails to produce antidiuretic hormone. If a person suffering from diabetes insipidus were to consume large amounts of a sugar-containing beverage, which of the following would most likely be observed after one hour? A) Production of concentrated urine B) Production of dilute urine C) Elevated plasma glucose levels D) Elevated urine glucose levels

B. An individual who cannot make ADH will be unable to reabsorb water from the collecting duct of the nephron, resulting in urine that is very dilute (choice B is correct, and choice A is wrong). Since diabetes insipidus is strictly a disorder involving ADH production, no effect would be seen on plasma glucose levels (very low after one hour due to the effects of insulin) or urine glucose levels, which should always be zero in a normal individual. Do not confuse this disorder with diabetes mellitus, a disorder involving insulin in which blood and urine glucose levels are abnormally high (choices C and D are wrong).

A woman who is 20 weeks pregnant presents for an amniocentesis. With a needle and ultrasound guidance, the physician withdraws some of the amniotic fluid for genetic analysis. A karyotype analysis performed on the amniotic fluid reveals an extra copy of chromosome 21. Which of the following is true of the child after birth? A) Kidney function will be impaired since the kidneys are underdeveloped. B) Digestion will be impaired since there is no communication between the small and large intestines. C) Sensory neural input will be diminished since there is no development of the sensory cortex. D) Muscle tone will be greater than expected.

B. As per the passage, the child with trisomy 21 may suffer from a prematurely ending small intestine (a blind-ending jejunum), and therefore, digestion will be impaired if not surgically corrected. Kidney dysfunction is not listed as a condition in trisomy 21 (choice A can be eliminated) nor is an underdeveloped sensory cortex (choice C can be eliminated). The passage says hypotonia is involved, which is a decrease in muscle tone, not an increase (choice D can be eliminated).

Excess calcitonin in the blood would most likely result in which of the following abnormalities? A) Abnormally weak bones B) Abnormally dense bones C) A greater than normal number of osteoclasts D) A decrease in the collagen content in bones

B. Calcitonin reduces blood calcium levels through three processes: decreasing calcium reabsorption by the kidney, decreasing calcium absorption by the small intestine, and activating osteoblasts, which store calcium as bone. This last function increases bone density (A is wrong, and B is correct). Osteoclasts degrade bone and are stimulated by parathyroid hormone, not calcitonin (C is wrong). Decreasing the collagen content in bones would also result in weaker bones (D is wrong), and in any case, the amount of collagen is not dependent on calcium or its regulatory hormones.

Novobiocin is an antibiotic that inhibits bacterial DNA gyrase. Its antibiotic properties are based on the fact that it: A) prevents DNA helix formation. B) inhibits DNA replication. C) prevents RNA transcription. D) inhibits protein translation.

B. DNA gyrase supercoils bacterial DNA (choice B is correct). It does not affect helix formation; the DNA is already in a double-helix before it is supercoiled (choice A is wrong). It does not affect protein synthesis, since that has to do with RNA and ribosomes, not DNA (choice D is wrong). The job of DNA gyrase is to continually introduce negative supercoils into the circular bacterial genome. Then, when the helix is opened for replication and positive supercoils are introduced (the DNA gets more tightly wound at the ends of the replication fork), the already-present negative supercoils cancel out the newly introduced positive supercoils, and DNA tension is kept to a minimum. If this does not occur, the DNA strand would become too tightly wound for replication to continue, and in the absence of replication, the bacteria cannot reproduce. Note that the same could be argued for transcription, but since that involves such a small region of the genome, the positive supercoiling is not a significant effect (choice C is wrong).

All of the following are components of plasma EXCEPT: A) fibrinogen. B) erythrocytes. C) potassium. D) immunoglobulins.

B. Plasma is the cell-free portion of blood. Proteins and salts are present in plasma, but cells—such as erythrocytes— are not, by definition.

Which of the following is true based on the chart below? Phospholipid Head group Phosphatidic acid Phosphate Phosphatidylethanolamine Phosphate and ethanolamine Phosphatidylcholine Phosphate and choline Phosphatidylserine Phosphate and serine A) Lipid and protein macromolecules can both contain serine, while lipids and nucleic acids have no biological building blocks in common. B) Components of the plasma membrane can contain nitrogen atoms and amino acids. C) Phospholipids are amphoteric, in that they have hydrophilic and hydrophobic properties or regions. D) At least some phospholipids contain three fatty acids connected to glyercol via ester linkages.

B. Proteins and phosphatidylserine both contain serine, and phospholipids and nucleic acids both contain phosphate groups (eliminate choice A). Based on the chart in the question stem, choice B is correct. Phospholipids can contain nitrogen atoms (as in ethanolamine and serine) and also amino acids (as in serine). Lipids are amphipathic because they contain both hydrophilic and hydrophobic regions. Amphoteric means that a molecule can act as either an acid or a base (eliminate choice C). Phospholipids contain a backbone (usually glycerol) with two fatty acids and a phosphate group. The phosphate group can be linked to other polar groups. Lipids that contain a glycerol backbone and three fatty acids are called triacylglycerides (eliminate choice D).

A woman affected by DM2 has two female children, the first child carries 3 additional CCTG repeats in her copy of the ZNF9 gene, while the second child carries 2 additional CCTG repeats in her copy of the ZNF9 gene. Which child would you expect to have the more significant symptoms? Please choose from one of the following options. A. The second child, as she has one less CCTG repeat. B. The first child, as she has a greater likelihood of having an out-of-frame gene. C. The second child, as she has a greater likelihood of having an out-of-frame gene. D. Without information regarding the mother's genes, there is not enough information to answer the question.

he severity of DM2 is caused by an increased number of CCTG repeats. Hint #2 We do not know whether the mother's ZNF9 gene matches the reading frame of the original gene. Hint #3 Since we do not know the reading frame of the mother's gene, it is not possible to predict if 2 or 3 repeat additions will result in a frame-shift. Therefore we cannot answer the question.

What would be the most effective negative control for the experiment described above? Please choose from one of the following options. A. Extract from keratinocytes that had been killed with UV light B. Extract from melanocytes post-UV exposure C. Extract from keratinocytes that had NOT been treated with UV light D. A known sample of POMC mRNA

his question requires the examinee to know and apply the definition of a negative control. In general, a negative control is designed such that no result is expected to occur. Hint #2 An effective negative control addresses the experimental hypothesis. In this case the hypothesis can be inferred to be that UV exposure influences POMC expression in keratinocytes. Hint #3 If the hypothesis concerns UV dependent transcription of POMC , then one must first know about UV independent transcription of POMC. A suitable negative control would be an extract from keratinocytes that had not been treated with UV light.

As described in the passage, which of the following best describes an evolutionary explanation for the amount of diversity present at HLA loci? A) The presence of different HLA alleles increases the fitness of individuals by decreasing susceptibility to infection. In an environment with continued exposure to pathogens, variation is then selected for, not against. B) New mutations at HLA loci decrease the fitness of individuals and are selected against, preventing new alleles from appearing within a population. C) HLA loci are under divergent selection, selecting against extremes at these loci. This increases the heterozygosity in the population. D) New mutations at HLA loci may increase resistance to rare or exotic pathogens. This increases the fitness of individuals, even in the absence of exposure to rare and exotic pathogens.

A. While nearly all of these answer choices may sound appealing by using terms associated with evolution and selection, it is important to note that the question states "as described in the passage." The only evolutionary explanation for increased diversity at HLA loci that is given in the passage is stated at the end of the second paragraph. The passage states that different HLA alleles may increase resistance (or decrease susceptibility) to infection. This would be an example of increased fitness (choice A is correct). Choice B can be eliminated because the statement is inaccurate. The passage states that variation increases fitness (by decreasing susceptibility to infection). Choice C can also be eliminated. While some HLA loci may be under divergent selection, variation does not necessarily imply heterozygosity (and neither are discussed in the passage, even implicitly). While choice D accurately states that mutations may increase resistance to infection, it states that only infection by rare or exotic pathogens is decreased. The type of infection is not discussed in the passage (choice A is better than choice D).

What might an inhibition of adenylyl cyclase cause? Please choose from one of the following options. Decreased effects of serotonin. Decreased effects of adrenaline. Increased effects of serotonin. Increased effects of adrenaline.

Adenylyl cyclase, also known as adenylate cyclase, produces the second messenger cAMP in regulating the effects of adrenaline. Hint #2 Morphine is known to be a painkiller, causing things like sleepiness, slowing our nervous system and blocking pain. Hint #3 Adenylyl cyclase produces cAMP from ATP which then causes the effects of adrenaline in our body, such as increased heart rate, dilation of skeletal blood vessels, and breakdown of glycogen into glucose for immediate energy. Morphine has opposite effects and inhibits adenylyl cyclase.

HIV, the virus that causes AIDS, is a retrovirus. Retroviruses have an RNA genome that, after entry into the host cell, is reverse transcribed to DNA, then incorporated into the host genome. RNA copies of the viral genome are produced using the normal host machinery, then packaged into viral capsid proteins for release. Which one of the following is most likely to contribute to the development of drug-resistant HIV? A) Mutation of the virus after insertion into the host-cell genome B) Frequent random errors in transcription by host-cell enyzmes C) Viral proteins folding differently in the presence of drug than in its absence D) Changes in the tertiary structure of viral RNA by drug

B. Drug resistance develops as mutant versions of the virus are produced and begin infecting new host cells. The change must be able to be passed on to viral progeny. If the viral proteins fold differently in the presence of drug than in its absence, this might account for the mechanism of action of the drug, but does not explain how drug resistance develops, nor could this be a heritable change (choice C is wrong). Changes in the tertiary structure of the viral RNA by the drug is another temporary, non-heritable change (choice D is wrong). Choices A and B are both plausible mechanisms for creating new, heritable versions of the virus. However, mutation of the virus after insertion into the host-cell genome would require that an error in DNA replication be made—a rather unlikely event, given the proofreading ability of DNA polymerase. Viral progeny are produced through transcription (creating a new RNA viral genome off the permanent DNA version inserted into the host-cell genome), and the question states that this occurs using the normal host machinery. Since the normal host RNA polymerases have no proofreading function, it is more likely that errors will occur here, leading to mutant virus (choice B is better than choice A).

It has been suggested that one reason tumor cells have novel metabolism is to generate biomolecules required to increase biomass, which is required when proliferation rates are high. How could this occur? A) Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis. B) Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides. C) High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism. D) Glutamate is converted into glutamine to provide the cell with an amino acid precursor, thus powering translation.

B. Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong), although this does support nucleotide and fatty acid biosynthesis. Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis (choice B is correct). The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). Paragraph 2 describes how glutamine is first converted into glutamate, not the other way around (choice D is wrong).

Which of the following is expected to be true of children with Down syndrome? A) Aortic arterial blood carbon dioxide concentration is normal. B) Aortic arterial blood oxygen saturation is lower than normal. C) Pulmonary arterial blood pressure is lower than normal. D) Pulmonary arterial blood oxygen saturation is lower than normal.

B. In children with Down syndrome, one of the primary cardiac defects is a truncus arteriosus, which presents as a common arterial trunk coming off both the left and right ventricles. This allows for mixing of blood from the left and right circulations. Since the right circulation is relatively high in carbon dioxide and the left circulation is relatively low in carbon dioxide, the resultant mix will be somewhere between (i.e., not normal, choice A is wrong). By similar reasoning, the right circulation that is typically low in oxygen will mix with the left circulation that is typically higher in oxygen, so the overall aortic arterial oxygen saturation will be less than normal (choice B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (choice D is wrong). Since a common arterial trunk is receiving blood from both the right and left ventricles, the pulmonary artery will be receiving more blood than usual and therefore will have a higher than normal blood pressure (choice C is wrong).

Which one of the following mutations would be most likely to convert a proto-oncogene into an oncogene? A) Silent mutation B) Missense mutation C) Nonsense mutation D) Deletion mutation

B. In order for the mutation to have the described effect, it must modify the protein without completely eliminating it or destroying its effect. A missense mutation converts a codon for one amino acid into a codon for a new amino acid, resulting in a small change within the protein's primary sequence, and an alteration (but usually not a total elimination) of the protein's function. A silent mutation converts a codon for an amino acid into a new codon for the same amino acid and has no effect on the protein product (A is wrong). A nonsense mutation creates a stop codon out of an amino acid codon, resulting in truncation of the protein and (usually) a loss of its function (C is eliminated). A deletion mutation eliminates one or more base pairs, altering the reading frame and drastically changing the amino acid sequence of the protein (and therefore eliminating its function, D is wrong).

A diethyl ether solution of benzaldehyde containing benzoic acid impurities may be purified by extraction with which of the following solutions? I. 10% NaHCO3(aq) II. 1.0 M HCl(aq) III. 1.0 M NaOH(aq) A) I only B) I and III only C) II and III only D) I, II, and III

B. In this Roman numeral question, eliminating choices that do not show differential solubility for benzaldehyde and benzoic acid is the best approach. Both benzaldehyde and benzoic acid are water-insoluble, so extractions that do not appreciably change the structure of one of the two will not allow effective separation. Because benzoic acid is acidic, it is deprotonated by NaHCO3, yielding the water soluble salt sodium benzoate. Since a salt is soluble in aqueous environments, item I is true (eliminate choice C). Neither benzaldehyde nor benzoic acid will react with or show differential solubility in the presence of the strong acid HCl, so both will remain dissolved in diethyl ether (eliminate choice D). In a similar fashion to using the weak base above, extraction with aqueous NaOH will induce salt formation by deprotonating the carboxylic acid (eliminate choice A).

In an experiment, various concentations of streptomycin were added to a culture of E. coli. At low concentrations of streptomycin, increased amount of misreading of the mRNA was observed. At high streptomycin concentrations, the initiation of protein translation was inhibited. Which of the following statements is consistent with these results? A) Streptomycin inhibits the large subunit (50S) of the ribosome. B) Streptomycin inhibits the small subunit (30S) of the ribosome. C) Streptomycin binds to RNA polymerase and blocks its function. D) Streptomycin binds to the mRNA and targets it for degradation.

B. Since the errors are observed in translation and not in transcription, it is likely the ribosome that is being affected and not RNA polymerase (choice C is wrong). The role of small subunit (30S) of ribosome is to initiate translation by recognizing the first AUG start site and recruiting the tRNAfmet and the large subunit. It is also important for proofreading and maintaining fidelity during the translation process. Therefore, inhibition of small subunit would result in decreased initiation of protein translation and mRNA misreading (choice A is wrong and choice B is correct). If streptomycin bound to mRNA and caused its degradation, we would likely see less protein being made at low concentration (due to less mRNA available), not misreading of the mRNA (choice D is wrong).

During spermatogenesis, spermatids: A) are frozen in meiosis II until after fertilization. B) have already undergone meiotic recombination. C) have four copies of the genome per cell. D) have no nucleus.

B. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (A is wrong). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (B is true). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (C is wrong). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (D is wrong)

Why is there a latent period in the production of antibodies during the primary response? A) The pathogen must first proliferate before the immune system can mount a response. B) Time is required for B lymphocytes to proliferate. C) Antibodies must be produced by T lymphocytes. D) The antigen must infect B lymphocytes prior to the production of antibodies.

B. The passage describes the latent period as the time required for the lymphocytes to begin to divide and differentiate into plasma and memory cells (choice B is correct). The pathogen need not proliferate for an immune response to be mounted; it only needs to be present (choice A is wrong). Antibodies are produced by B lymphocytes, not T lymphocytes (choice C is wrong), and an antigen does not infect the B cells, it must only bind to them (choice D is wrong).

A person who suffers partial destruction of the lateral lemniscus on the left side would be expected to suffer which of the following? A) Ipsilateral hearing loss B) Bilateral hearing loss C) Total sensorineural deafness D) Defects in auditory processing and association

B. The passage states that the first decussation of the auditory pathway occurs at the connection between the cochlear nucleus and the superior olivary nucleus. Thus, any lesion that occurs past that point will result in binaural, rather than monaural, hearing loss (choice A is incorrect). By the same token, total sensorineural deafness would require bilateral destruction of the hearing pathway (choice C is incorrect). Auditory processing and association occur in the higher brain, after processing by the primary auditory cortex (choice D is incorrect). Bilateral hearing loss (choice B) is correct; a lesion on one side of the brain will result in partial but not complete hearing loss, since auditory information ascends bilaterally.

When conducting a hearing test, sounds of varying pitches and loudness are created in order to determine if any aspect of hearing has been compromised. Which of the following best describes how differences in pitch are determined by the brain? A) Differences in pitch are determined by the amplitude of the vibration of the basilar membrane. Larger vibrations cause more frequent action potentials and smaller vibrations cause less frequent action potentials. B) Differences in pitch are determined by which region of the basilar membrane vibrates, stimulating different auditory hair cells. High pitched sounds vibrate the basilar membrane near the oval window and low pitched sounds vibrate the basilar membrane farthest from the oval window. C) Differences in pitch are determined by the amplitude of the vibration of the basilar membrane. Smaller vibrations cause more frequent action potentials and larger vibrations cause less frequent action potentials. D) Differences in pitch are determined by which region of the basilar membrane vibrates, stimulating different auditory hair cells. High pitched sounds vibrate the basilar membrane furthest from the oval window and low pitched sounds vibrate the basilar membrane near the oval window.

B. This is a 2x2 question, where one piece of information can eliminate two answer choices. Differences in pitch are determined by the region of the basilar membrane that vibrates, and loudness is determined by amplitude of the vibration (choices A and C are wrong). High pitched sounds vibrate the region of the basilar membrane near the oval window, and low pitched sounds vibrate the basilar membrane farthest from the oval window (choice B is correct and choice D is wrong).

During a mastectomy, the lymph nodes are often removed in addition to the tumor. This surgical procedure will most likely: A) render the patient less susceptible to viral infections. B) stop the spread of breast cancer to the rest of the body. C) increase the number of circulating lymphocytes. D) remove the oncogenes within the affected tissue.

B. When cancers metastasize, tumor cells leave their original site and enter the blood or lymph vessels to travel to a new part of the body where they can implant and grow. Therefore, removal of nearby lymph nodes during surgery to remove a cancerous tumor is often done to help ensure that the cancer cannot spread (choice B is correct). Removal of the lymph nodes could render the patient more susceptible to infection and decrease the number of circulating lymphocytes (choices A and C are wrong). Finally, to remove the oncogenes within the affected tissue, the surgeons would only need to remove the breast tissue since the lymph nodes are not affected (choice D is wrong).

Tryptophan, an essential amino acid found in banana, turkey, and milk proteins, can induce sleep in some people. Warm milk causes greater sleepiness than cold milk because heating the milk: A) reduces the solubility of tryptophan in the milk. B) causes hydrolysis of lactose, releasing tryptophan. C) releases free tryptophan from proteins, causing more rapid intestinal absorption. D) increases the rate of absorption of tryptophan by the stomach.

C. Choice A does cannot be correct: if heating the milk reduced tryptophan solubility, this would decrease, not increase the sleep-inducing properties of milk. Choice B is wrong: lactose is a disaccharide, not a protein, and its hydrolysis cannot release an amino acid. Choice D cannot be correct: amino acids, like most nutrients, are absorbed in the small intestine mostly, not the stomach. The best answer is C. Although heating the milk does not create more tryptophan, it might help to hydrolyze some of the milk proteins and release tryptophan so it can be absorbed more rapidly after ingestion and cause greater sleepiness.

Which of the following could explain the Warburg effect? I. Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. II. Cells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. III. Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death. A) I only B) III only C) I and III only D) I, II, and III

C. Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).

Which of the following represents the major products of the saponification of linolein, a triacylglycerol in which glycerol is esterified with linoleic acid? A) 1 Glycerol and 1 linoleic acid B) 3 Glycerol and 1 potassium linoleate C) 1 Glycerol and 3 potassium linoleate D) 1 Glycerol and 3 linoleic acid

C. A triacylglycerol is composed of a glycerol backbone and three fatty acids, in this case three linoleic acid molecules. Saponification of linolein yields one glycerol molecule (eliminate choice B) and three linoleic acid salts, so three potassium linoleate equivalents are expected (eliminate choice A). Linoleic acid is not seen, as fatty acids do not remain protonated under basic conditions (eliminate choice D).

During peak cardiac activity, heart muscle utilizes the anaerobic pathway for only 10% of its energy as opposed to up to 85% for maximally contracting skeletal muscle. With respect to skeletal muscle, cardiac muscle would: A) convert more pyruvate to lactate. B) consume 1/8 as much ATP to produce the same contractile force. C) consume less glucose per ATP produced. D) require less oxygen per molecule of ATP produced.

C. Aerobic respiration includes glycolysis, the PDC, the Krebs cycle, and electron transport, to produce a total of 32 ATP per glucose (assuming the malate-aspartate shuttle is used to transfer the electrons from glycolytic NADH into the mitochondria). By contrast, anaerobic respiration involves only glycolysis and fermentation of pyruvate to lactate or alcohol, to produce 2 ATP per glucose. Anaerobic respiration is therefore much less efficient and must consume much more glucose to produce the same amount of ATP. Since cardiac muscle uses mostly aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct). Conversion of pyruvate to lactate occurs during fermentation, which cardiac muscle carries out less of, not more (choice A is wrong). The amount of contractile force generated per ATP is independent of the source of ATP used (choice B is wrong). Cardiac muscle uses more aerobic respiration, so it will use more oxygen per ATP produced, not less (choice D is wrong).

Lysine is an amino acid with a basic side chain. Its isoelectric point would be closest to: A) 6. B) 7. C) 8. D) 12.

C. All amino acids have one low pKa ~2 (for the acidic carboxyl group) and one high pKa ~10 (for the basic ammonium group), which are averaged to calculate the pI. When an acidic or basic side chain contributes a third pKa value, the pI is determined by averaging the two lowest or the two highest pKa values of the three. Thus, a basic side chain, such as Lys, will contribute a high (~10) third pKa value and is expected to increase the pI to a point above neutral pH (eliminate choices A and B). Choice D is not possible as a pI of 12 implies an average of two very high pKa values, which are not found in amino acids.

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true? A) Yes; bacteria are capable of undergoing genetic recombination through a variety of mechanisms. B) Yes; bacteria reproduce sexually, and the progeny of any one cell are genetically distinct from the parent cell. C) No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. D) No; bacteria can reproduce only by meiosis, which ensures preservation of the genome.

C. Bacteria reproduce asexually, by one cell replicating its genome and then splitting into two cells that are genetically identical to the original cell. The only potential sources of genetic variation in bacteria are mutation and the transfer of genetic information through conjugation, transduction, or transformation, none of which are linked to reproduction. In the absence of mutation, all progeny of a cell will be identical to the original cell (C is true). Bacteria only perform recombination under special circumstances such as through the presence of Hfr plasmids that replicate a portion of the bacterial genome to make it transiently diploid (A is false). There is no indication of a role for Hfr in this case and in a clonal cell line, it could not play a role. Bacteria do not perform the recombination, independent assortment and independent segregation that create genetic diversity in eukaryotes that reproduce sexually (B is wrong). They also do not perform meiosis (D is wrong).

The most likely explanation for the inverse relationship between mammal size and pulse is that: A) most small mammals live under anaerobic conditions. B) large animals tend to consume proportionately more food than smaller animals. C) smaller mammals have a higher metabolic rate for a given mass of tissue than larger mammals. D) smaller mammals have reduced oxygen requirements for a given mass of tissue.

C. Choice A is wrong since all mammals use primarily aerobic respiration except for short periods of unusually strenuous activity. Choice B is wrong since it does not directly involve metabolic rate or pulse and would predict that the metabolic rate and pulse rate will remain the same despite size. Choice D is wrong since reduced oxygen requirements would predict a slower pulse. Choice C is the best choice: If tissues in smaller animals have a higher metabolic rate, they will require a greater supply of blood to provide oxygen and nutrients and carry away carbon dioxide. To meet these greater needs, the pulse rate could be increased.

Fatty acid synthase is most likely suppressed by all of the following hormones EXCEPT: A) glucagon. B) thyroid hormone. C) insulin. D) cortisol.

C. Fatty acid synthesis will only be active if the cells have enough energy and are ready to build storage molecules, especially long term storage molecules. Insulin is released in response to elevated blood glucose and the glucose can be used for cellular respiration and storage (choice C is correct). For a process of elimination approach: glucagon is a hormone that liberates glucose from glycogen, raises blood glucose levels, and induces fat/triglyceride breakdown (eliminate choice A). Thyroid hormone increases the basal metabolic rate of cells. This will increase the ATP consumption of the body and limit production of any long-term storage molecules (eliminate choice B). Cortisol is a stress hormone and will lead to lipolysis to increase ATP availability in stressful situations. It will inhibit lipogenesis (eliminate choice D).

Which of the following best describes the role of fructose-2,6-bisphosphate? A) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating fructose-1,6-bisphosphatase and inhibiting phosphofructokinase. B) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting hexokinase. C) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase. D) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating hexokinase and inhibiting fructose-1,6-bisphosphatase.

C. Fructose-2,6-bisphosphate exerts reciprocal control on glycolysis and gluconeogenesis through phosphofructokinase and fructose-1,6-bisphosphatase, and has no impact on hexokinase activity (choices B and D are wrong). Fructose-2,6-bisphosphate stimulates phosphofructokinase (glycolysis) and inhibits fructose-1,6-bisphosphatase (gluconeogenesis). Choice C is correct, and choice D can be eliminated.

Glucose-6-phosophate dehydrogenase (G6PDH) deficiency is an inheritable metabolic disorder that can result in the destruction of red blood cells, and in severe cases, can lead to kidney and liver failure. Which of the following best describes the reason why G6PDH deficiency can result in premature red blood cell destruction? A) G6PDH deficiency results in a buildup of glucose-6-phosphate, increasing the solute concentration in the intracellular environment, causing the cell to swell and lyse. B) G6PDH deficient cells are recognized as abnormal and are targeted for destruction by macrophages. C) G6PDH deficiency results in a deficiency of NADPH, which results in increased damage from reactive oxygen species and causes premature cell death. D) G6PDH deficiency results in an increased intracellular pH, which denatures proteins and triggers apoptosis.

C. Glucose-6-phosphate dehydrogenase (G6DPH) catalyzes the first step in the pentose phosphate pathway, where NADPH is produced. NADPH plays a major role in preventing damage due to oxidative stress/reactive oxygen species. A deficiency of G6PDH prevents the production of NADPH, thus individuals with this deficiency would experience increased damage from reactive oxygen species (choice C is correct). G6PDH deficiency would not be expected to alter the intracellular osmolarity (choice A is incorrect) or to increase the intracellular pH (choice D is also incorrect). Choice B is incorrect: macrophages are not involved in the death of red blood cells in G6PDH cells.

Based on the results of Experiments 1 and 2, it is reasonable to assume that all components of the mechanism necessary for concentrating lactose intracellularly within E. coli are located in the: A) rough endoplasmic reticulum. B) genome. C) cell membrane. D) Golgi bodies.

C. Here, the key experiment is Experiment 2, in which vesicles formed from the cell membrane had the same ability to concentrate lactose as whole cells (C is correct). The genome will encode protein components, but the active components are not themselves physically located in the genome (B is wrong). The ER and Golgi are not found in bacteria (A and D are wrong).

In gastroesophageal reflux disease (GERD), gastric secretions exit the stomach via the lower esophageal sphincter and come in contact with the lower part of the esophagus. How would this affect the esophageal epithelium? Increased proliferation of the stratified squamous epithelium Conversion to columnar epithelium similar to that in the stomach Introduction of bicarbonate-secreting cells A) I only B) II only C) I and II only D) I, II, and III

C. Item I is true: in the short term, gastric acid would corrode the outermost layer of epithelium in the lower esophagus, increasing the production of cells (choice B can be eliminated). Item II is true: over time, the acid will destroy the epithelium and expose the submucosa. Columnar epithelium found in the stomach will gradually encroach on this area and become the dominant form of epithelium, since it is not sensitive to acid (choice A can be eliminated). Item III is false: There are specialized areas known as Brunner's glands in the proximal duodenum that produce bicarbonate to neutralize the acidic chyme as it passes from the stomach to the duodenum. However, there is no reason to assume that these cells would traverse the stomach to the lower esophagus (choice D can be eliminated and choice C is correct).

The function of galactoside permease is to transport lactose across the bacterial cell membrane. Galactoside permease is necessary because: A) lactose is hydrophobic and cannot cross the lipid bilayer. B) lactose needs a protein carrier to move it against its concentration gradient. C) lactose is hydrophilic and cannot cross the lipid bilayer. D) lactose needs to be converted to allolactose to enter the cell.

C. Lactose, like other carbohydrates, is very hydrophilic, not hydrophobic (choice A is wrong). Hydrophilic molecules are unable to diffuse across a membrane into the cell (choice C is correct). There is no information to support B; in any case, it can be assumed that lactose is not being transported against a gradient (choice B is wrong). Allolactose only signals the presence of lactose, but it is lactose that is the energy source, and it is lactose that is imported into the cell (choice D is wrong).

Which of the following best describes the cascade that leads to cellular apoptosis after exposure to intra- or extracellular death signals? A) Effector kinases cluster together and activate each other. The activation of effector kinases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. B) Effector caspases cluster together and activate each other. The activation of effector caspases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. C) Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. D) Effector kinases cluster together with initiator capsases, activating the initiator capsases. The activation of effector kinases leads to the deactivation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.

C. This is a recall question pertinent to concepts discussed in the passage (apoptosis). Apoptosis is carried out by proteases called caspases, which cleave targets at aspartic acid residues. This eliminated choices A and D, which both indicate that apoptosis is started by effector kinases. Choice C is correct: Initiator caspases cluster together and initiate apoptosis (thus the name). These initiator caspases can then activate effector caspases, which cleave proteins at aspartic acid residues; this triggers apoptosis. Choice B is incorrect, and confuses the role of initiator and effector caspases.

If radiolabeled stilbestrol were administered to the experimental chicks, stilbestrol would be found most heavily concentrated: A) at the cell membrane of oviduct tissue. B) in the cytoplasm of oviduct tissue. C) in the nuclei of oviduct tissue. D) in the mitochondria of oviduct tissue.

C. When stimulated by the addition of a ligand such as stilbestrol, estrogen receptor will localize within the nucleus, where it regulates genes by binding to enhancers and promoters. Radiolabeled stilbestrol would localize with estrogen receptor in the nucleus (C is correct). There is no estrogen receptor in the plasma membrane or mitochondria (A and D are wrong). Some estrogen receptor may be located in the cytoplasm, particularly in the absence of ligand, but it will localize mostly in the nucleus when it has ligand bound (B is wrong).

What do you expect to observe about the flow of electrons in the electron transport chain in a catabolic state vs. an anabolic state? Please choose from one of the following options. A. The flow of electrons will be faster in the catabolic state than the anabolic state. B. The flow of electrons will be the same in the anabolic state and the catabolic state. C. The flow of electrons will be faster in the anabolic state than the catabolic state. D. The flow of electrons will be different in the anabolic state and the catabolic state, but it is not possible to know whether one will be increased or decreased.

Catabolism describes energy-releasing reactions and anabolism describes energy-requiring reactions of metabolism. Hint #2 ATP synthesis is coupled to the proton gradient that is produced by the electron transport chain. Hint #3 By Le Chatelier's principle, an increase in ATP hydrolysis in the reactions of anabolism will increase the amount of ADP available for the ATP synthase. An increase in ADP substrate will, by Le Chatelier's Principle again, "push" the phosphorylation of ADP forward, requiring that the flow of electrons increase. The reverse is true in the catabolic state in which ATP is largely being produced, which will, by Le Chatelier's principle, move the reaction discussed above in the opposite direction

According to the Hardy-Weinberg equation, what is the expected frequency of the HLA-A2 and HLA-A3 alleles in the next generation (F3)? A) The expected frequency of the HLA-A2 allele is 0.9, and the expected frequency of the HLA-A3 allele is 0.1. B) The expected frequency of the HLA-A2 allele is 0.7, and the expected frequency of the HLA-A3 allele is 0.3. C) The expected frequency of the HLA-A2 allele is 0.6, and the expected frequency of the HLA-A3 allele is 0.4. D) Allele frequencies for the F3 generation cannot be predicted using Hardy-Weinberg equations because this population is not in Hardy-Weinberg equilibrium.

D. According to Table 1, the allele frequencies in this population are changing over time. Thus, Hardy-Weinberg does not apply, and the equations associated with Hardy-Weinberg cannot be used to predict allele frequencies in the next generation (choice D is correct, and all others can be eliminated).

If a patient with cystic fibrosis receives a double-lung transplant from a non-cystic fibrosis donor, would the new lungs be expected to develop cystic fibrosis? A) Yes, once you have cystic fibrosis, it develops in every organ of the body. B) Yes, since the primary defect is with respiratory secretions. C) No, because the infectious causes of the disease will be removed when the old lungs are taken out. D) No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene.

D. Cystic fibrosis is a genetic disease based on the abnormal protein CFTR. In a set of lungs from a person without cystic fibrosis, the lung cells presumably have normal-functioning CFTRs. Therefore, the new lungs should not be subject to the development of cystic fibrosis (eliminate choices A and B). Cystic fibrosis is a multi-organ disease since the CFTR is used in secretions from several glands, but this does not cause normal CFTRs to become abnormal. The primary defect is the protein CFTR, not the pulmonary secretions or an infectious cause (choice C is incorrect and choice D is correct).

During times of starvation: A) triacylglyceride stores in adipocytes are anabolized by lipase. B) fatty acids are broken down via oxidation, hydration and cleavage reactions, to generate NADPH, and to oxidize FADH2. C) glycerol is converted to DHAP, a glycolysis intermediate, which is then used to generate lactate and ethanol via fermentation. D) fatty acids are transported in the blood via binding to blood proteins like albumin, then are broken down via β-oxidation in target tissues.

D. During times of starvation, sugar, protein and fat stores are degraded to provide the body with sufficient energy. Triacylglycerides are catabolized (not anabolized) in adipocytes via lipase (eliminate choice A). The glycerol backbone is converted to DHAP, a glycolysis intermediate, which is then converted to pyruvate and put through Krebs cycle and ETC to make energy, or is used in gluconeogenesis to make glucose-6-phosphate. Fermentation can generate lactate or ethanol, but not both, and is only done in anaerobic conditions (eliminate choice C). Fatty acids are transported in the blood via binding to blood proteins like albumin. Once at the target tissue, they are activated, shuttled into the mitochondria and broken down via β-oxidation. This generates acetyl-CoA (via cleavage of the carbon backbone), as well as NADH and FADH2 (via reduction of NAD+ and FAD respectively, eliminate choice B and choice D is correct). Note that NADPH is made via the pentose phosphate pathway, not during fatty acid oxidation.

Which of the following represents the correct sequence for embryogenesis? A) Fertilization → gastrulation → blastulation → neural tube formation → somite formation B) Fertilization → gastrulation → blastulation → somite formation → neural tube formation C) Fertilization → blastulation → neural tube formation → gastrulation → somite formation D) Fertilization → blastulation → gastrulation → neural tube formation → somite formation

D. Fertilization is the first step, followed by a series of rapid cell cleavages to form a hollow ball of cells called the blastulam (eliminate choices A and B). Next comes the gastrula, in which cells move into the interior of embryo to form the three germ layers (eliminate choice C). Gastrulation is followed by the formation of the neural tube, which will form the nervous system, followed by the formation of other organs and tissues, such as the somites that will differentiate into bones and muscle. This makes choice D the only possible correct order of events.

Vinca alkaloids are a class of anti-cancer drugs derived from the periwinkle plant. Once absorbed into a cell, they interfere with the polymerization of microtubules. These drugs can prevent cancer from spreading by disrupting: A) pseudopod formation, thereby preventing cellular locomotion. B) prophase, thereby halting mitosis. C) transcription, thereby halting production of crucial cell proteins. D) metaphase, thereby halting tumor growth.

D. Formation of microtubules is crucial for formation of the metaphase plate. Microtubules polymerize from the centrioles outward. They contact centromeres (to become kinetochore fibers) and "push" the chromosomes towards the center of the cell to form the metaphase plate. Without proper microtubule polymerization, metaphase (and the rest of mitosis) cannot occur. If mitosis cannot occur, then neither can tumor growth. Note that microtubules are also required during prophase in forming the mitotic spindle; however, some parts of prophase can still occur in the absence of microtubule formation (DNA condensation, loss of the nuclear membrane). This makes choice D a better choice than choice B, since virtually all of metaphase depends on proper microtubule polymerization. Pseudopod formation requires the growth of microfilaments (actin fibers), not microtubules (choice A is wrong), and transcription (RNA polymerization) does not require microtubules at all (choice C is wrong).

Based on Figure 1, which of the following is least supported? A) The OCA2 proteins in the eight species studies would be considered homologous. B) The OCA2 protein is conserved across eukaryotes. C) Zebrafish and Japanese killifish have a OCA2 common ancestor protein which is more related that that between horse and human. D) Humans and chimpanzees are more closely related than fission yeast and the wild boar.

D. Homologous proteins or genes are those that have evolved from a common ancestor. This matches the information in Figure 1, since the OCA2 protein in all organisms on the figure originated with the ancestor protein represented by the point at the bottom (choice A is supported and can be eliminated). The eight organisms on the figure are all eukaryotes and from diverse families (fungi, mammals, fish; choice B is supported and can be eliminated). Remember that yeast are fungi and therefore eukaryotic. OCA2 in zebrafish and Japanese killifish share a common ancestor which is not far away in evolutionary terms (i.e., is not very far down the diagram). In contrast, the common ancestor protein between human and horse OCA2 protein is the point at the bottom of the diagram; this is farther away in evolutionary terms (choice C is supported and can be eliminated). While choice D may be true based on logic and background information on evolution, it is not supported by Figure 1; this phylogenetic tree contains information on how the OCA2 proteins are evolutionarily related, not how organisms are related (choice D is the least supported and the correct answer choice).

It was observed under the microscope that a small section of bone consisted of concentric rings of fibrous material surrounding an open circular region. The bone section was identified as an osteon. The circular region is expected to contain: A) mature osteocytes. B) maturing chondrocytes. C) large blood cell precursors. D) blood vessels.

D. In compact bone, the osteon is the basic unit of structure. Concentric rings surround the central circular region that contains blood or lymph vessels (choice D is correct). Osteocytes can be found in small spaces along channels branching away from the central channel (choice A is wrong). Bone is the site of the marrow in which blood cell precursors are found, but bone marrow is found in spongy bone and not in compact bone where osteons are found (choice C is wrong). Chondrocytes are found in cartilage (choice B is wrong).

Thirteen amino acids, including methionine, valine and proline, are glucogenic in humans. This means their α-keto acid carbon skeleton is converted to pyruvate during amino acid catabolism. After deamination, valine can therefore: Be converted into CO2 and H2O to generate ATP. Generate at least three NADH and two FADH2. Enter gluconeogenesis to generate glucose. A) I only B) III only C) I and II only D) I and III only

D. Item I is true: Amino acids are catabolized via deamination into α-keto acids and ammonia. Based on the information in the question stem, the α-keto acid formed from valine will be converted to pyruvate. Pyruvate can keep going through cellular respiration to generate CO2, H2O and ATP. Eliminate choice B. Item II is false: Pyruvate is converted into one acetyl-CoA (during which 1 NADH is made), and the acetyl-CoA would then generate three NADH and only one FADH2 as it cycles through the Krebs cycle. Eliminate choice C. Item III is true: Pyruvate can also enter gluconeogenesis to generate glucose. Eliminate choice A and choice D is correct.

In a patient presenting with a mesothelioma-induced pleural effusion, which of the following would be expected to be increased? Pleural pressure Respiratory rate Heart rate A) I only B) I and II C) II and III D) I, II, and III

D. Item I is true: As fluid fills the pleural space, the pressure would become more positive (choice C can be eliminated). Item II is true: The increased pressure on the lungs makes expansion, and thus breathing, difficult. The patient will need to breathe more frequently in order to move the same amount of air (choice A can be eliminated). Item III is true: An increased heart rate will accompany the increase in respiratory rate in order to increase the amount of blood moving through the lungs. This would hopefully improve oxygenation (choice B can be eliminated and choice D is correct).

Which of the following would be a major concern of mortality in Myasthenia Gravis patients? A) Cardiac failure B) Muscle weakness in the upper limbs C) Paralysis of the facial muscles D) Respiratory distress

D. MG is a disease affecting skeletal muscle; therefore, all skeletal muscles in the body are susceptible (choice A can be eliminated). A very important skeletal muscle in humans is the diaphragm; in severe MG patients, death can occur due to respiratory distress or arrest. Paralysis of facial muscles and weakness of upper limbs can occur; however, they are generally not the primary cause of mortality (choices B and C can be eliminated). Choice D is correct.

A chemist prepares a 1 M solution of sulfuric acid. Which of the following gives the relative concentrations of the species in solution? A) [H2SO4] > [HSO4-] > [SO42-] B) [SO42-] > [HSO4-] > [H2SO4] C) [HSO4-] > [H2SO4] > [SO42-] D) [HSO4-] > [SO42-] > [H2SO4]

D. Since sulfuric acid is a strong acid and dissociation of the first proton is complete, undissociated H2SO4 will have the lowest relative concentration, eliminating choices A and C. Since the dissociation constant for the loss of the second proton is well below 1 (~ 10-2), HSO4- will exist in greater concentration than SO42-, making choice D the correct answer.

Exposure to asbestos may trigger scarring of the lungs, causing pleural fibrosis (development of fibrous tissue in the pleura), a subsequent loss of lung elasticity, and a reduction in total lung capacity. These are the hallmarks of restrictive lung disease. Shortness of breath in asbestosis would be MOST similar to shortness of breath due to: A) intense cardiovascular exercise. B) a low oxygen environment. C) inflamed airways during an asthma attack. D) diaphragm paralysis from a spinal cord injury

D. The key words in the question text are "reduction in total lung capacity" and "loss of elasticity." Although exercise and a low oxygen environment can result in shortness of breath, there is no reduction in total lung capacity and the lungs remain compliant and elastic (choices A and B can be eliminated). Asthma is more properly described as an obstructive disease, in which air flow is reduced due to narrowed bronchial tubes. Again, there is no loss of elasticity or reduction in total lung capacity (choice C can be eliminated). However, diaphragm paralysis would result in a significant reduction of the ability of the lungs to expand. Although intercostal muscles can attempt to compensate for the loss of diaphragmatic function, they cannot produce the size-changes in the chest cavity that the diaphragm can. Thus, lung expansion would be reduced and total lung capacity would fall. Medically this is classified as a "neuromuscular restrictive lung disorder (choice D is correct).

What type of cell is the origin of an acoustic neuroma? A) Microglia B) Astrocyte C) Oligodendrocyte D) Schwann cell

D. The passage states that an acoustic neuroma is a cancer of the myelin-producing cells of the vestibulocochlear nerve, a cranial nerve. Cranial nerves are part of the peripheral nervous system, so they are myelinated by Schwann cells (choice D) rather than oligodendrocytes (choice C is incorrect). Microglia (choice A) have an immune role in the CNS, while astrocytes (choice B) carry out various maintenance and regulatory activities in the CNS.

After an adult with cystic fibrosis ingests a carbohydrate-rich meal, which of the following would you expect to occur? A) High intracellular concentration of glucose B) Increased secretion of insulin C) Increased secretion of glucagon D) High extracellular concentration of glucose

D. The passage states that people with cystic fibrosis tend to have autodestruction of the pancreas. Autodestruction would eliminate the islets of Langerhans which secrete both glucagon and insulin (choices B and C are incorrect). After a carbohydrate-rich meal, the serum glucose concentration should rise (choice D is correct). In the absence of insulin, the serum glucose concentration will remain high until it can be cleared from the blood by the kidney (choice A is incorrect).

PCl5(g) leftwards harpoon over rightwards harpoon PCl3(g) + Cl2(g) ΔHo = 92.5 kJ/mol Two identical reaction vessels, X and Y, are charged with equal amounts of gaseous PCl5, and both trials are allowed to reach equilibrium. If flask X is held at 300 K while flask Y is maintained at 500 K, which of the following will be true about the reactions? A) Equilibrium is reached faster in flask Y because higher temperatures shift the reaction toward products. B) Equilibrium is reached faster in flask Y because ΔGY is more negative than ΔGX. C) Equilibrium is established at the same rate in both flasks because the K value of a reaction is a constant. D) Equilibrium is reached faster in flask Y due to the increased collision frequency of reactant molecules.

D. The speed that equilibrium is reached is based on the kinetics of the forward and reverse reactions and not on any thermodynamic parameters. Since both the forward and reverse reactions of the equilibrium will be sped up by increasing temperature, flask Y will reach equilibrium faster (eliminate choice C). The reason must be a kinetic principle, not a thermodynamic principle, so using ΔG or Le Châtelier's Principle, both thermodynamic concepts, as explanations must be wrong (eliminate choices A and B). According to the kinetic molecular theory, temperature is a measure of the average speed (velocity) of the particles. Increasing temperature increases speed, which in turn increases the frequency of collisions between the particles and the vessel.

All of the following tissues are part of the immune response EXCEPT the: A) thymus. B) spleen. C) lymph nodes. D) pancreas.

D. The thymus is the site of T-cell maturation (choice A can be eliminated), the spleen is the site of B-cell maturation and also filters antigens from blood (choice B is wrong), and lymph nodes are sites of both B- and T-cell proliferation (choice C can be eliminated). The pancreas produces digestive enzymes, bicarbonate, and hormones but is not involved in the immune response (choice D is not true and is thus the correct answer choice).

A biochemist determines reaction rate using varying substrate concentrations and two units of dihydrolipoyl transacetylase, the E2 enzymatic component of the pyruvate dehydrogenase complex, at standard state. The following data are obtained. What is the Km and Vmax of this enzyme under these conditions? A) Km = 17.5 µmol, Vmax = 35 µmol/min B) Km = 9.2 µmol, Vmax = 35 µmol/min C) Km = 17.5 µmol, Vmax = 44 µmol/min D) Km = 9.2 µmol, Vmax = 44 µmol/min

D. Vmax is the maximum reaction rate, or 44 µmol/min (eliminate choices A and B). Km is the amount of substrate required to get to ½Vmax (22 µmol/min). Based on data in the chart, Km will be a little below 10 µmol (eliminate choice C, choice D is correct).

What would you expect to happen to the fatty acid synthesis in patients with Glucose-6-Phosphate deficiency? Please choose from one of the following options. Increase Decrease Stay the same It depends

Decrease. NADPH is also an important source of "reducing power" for the synthesis of fatty acids. Hint #2 G6PD deficiency causes a decreased production of NADPH. Hint #3 Fatty acid synthesis will likely be halted since less NADPH is available to donate electrons in this process.

What is a plausible explanation for the boy's enlarged liver? Please choose from one of the following options. The liver contains an excess of glycogen The liver is engorged because of increased blood flow The liver contains an excess of glucose The liver grows faster than the rest of the body

Given the passage details, there is an abnormality in carbohydrate metabolism. What two processes are responsible for maintaining blood glucose levels? Hint #2 Glycogen and gluconeogenesis are responsible for maintaining blood glucose levels. Hint #3 The liver is one of the major stores of glycogen - normally, it is broken down after a meal to maintain blood glucose levels. If this is not occurring, it will build-up and cause the liver to enlarge.

Why do some patients with CPTII (unable to oxidize fatty acids) deficiency suffer from hypoglycemia? Please choose from one of the following options. Fatty-acid oxidation fuels gluconeogenesis Fatty-acid oxidation promotes insulin secretion Fatty-acid oxidation yields free glucose Fatty-acid oxidation yields Acetyl CoA

Hint #1 How does the body respond to a state of starvation? Hint #2 The body relies on its glycogen stores and then reverts to gluconeogenesis. However, gluconeogenesis is an anabolic process and cannot proceed without ATP. Hint #3 Fatty-acid oxidation fuels gluconeogenesis. Being unable to effectively oxidize fatty acids will compromise gluconeogenesis, which maintains blood glucose levels in a fast.

Fischer Projection

In a fisher projection, the horizontal substituents are coming out of the page, while those vertical substituents are going into the page.

The reaction of D-Glucose with an excess of methanol in the presence of an acid results in what product? Please choose from one of the following options. Hemiacetal Acetal Ketal Hemiketal

In the first step, the carbonyl carbon undergoes a nucleophilic attack by a hydroxyl group. Hint #2 Remember, a monosaccharide (like glucose) is more stable in the six membered ring form, so the first hydroxyl addition comes in the form of an intermolecular ring closure as the 5th carbon's hydroxyl group attacks the carbonyl carbon. This first step results in a hemiacetal formation. Hint #3 Acetal. The question stem states that the reaction occurs with excess alcohol reactants in an acidic environment. The hemiacetal is converted into an acetal.

Increasing the permeability of the inner mitochondrial membrane would have what effect on brown fat cells? Please choose from one of the following options. A. More heat production, More ATP production B. Less heat production, Less ATP production C. More heat production, ATP production unaffected D. Less heat production, ATP production unaffected

Increasing the permeability of the inner mitochondrial membrane will dissipate the proton gradient. Hint #2 Without a proton gradient, Le Chatelier's principle states that the flow of protons will be reduced through ATP Synthase and UCPs. Hint #3 Since the flow of protons is reduced through ATP Synthase and UCPs, there will be less heat production and less ATP production. Note that having more UCPs does not mean that ATP Synthase will not be affected, too, since both of these proteins rely on the proton gradient.

Given the enzymatic mechanism of NTKA deficiency, what are the most likely genotypes of patients 1 & 2? A. heterozygous & homozygous dominant B. homozygous dominant & heterozygous C. Homozygous recessive & homozygous dominant D. Heterozygous & homozygous recessive

It can be inferred from the plot in Figure 1 that the NTKAse from patient 1 & the NTKAse from patient have the same measured Vmax but different measured Km values (the y-intercept in the plot is equal to 1/Vmax and the x-intercept in the plot is equal to -1/Km). The NTKase from patient 1 has a lower measured Km value than the NTKAse from patient 2. A lowered measured Km value means the enzyme has a higher binding affinity for the substrate. From the passage, we know that NTKAse is hyperactive & binds TKA with much higher affinity than does wild-type NTKAse. A lower measured Km value of patient 1's NTKAse tells us they are homozygous recessive individual, having only NTKAse. Patient 2 could be homozygous dominant or heterozygous, either of which would have normally functioning NTKA. The genotypes of patients 1 & 2 are most likely to be homozygous recessive & homozygous dominant (or heterozygous), respectively.

It turns out that alcoholics, who often have limited glycogen stores, can also suffer from ketoacidosis. Of note, the enzyme used to metabolize ethanol also reduces a molecule of NAD+ to NADH. Why might alcoholics be susceptible to ketoacidosis? Please choose from one of the following options. A. Increased NADH levels increase flow of acetyl CoA through the Krebs Cycle B. Ethanol inhibits the hydrolysis of triacylglycerides in adipose tissue C. Increased NADH levels impairs gluconeogenesis D. Ethanol inhibits fatty-acid oxidation in adipose tissue

Ketoacidosis occurs as a result of increase ketone synthesis from acetyl CoA. Increased ketone synthesis is stimulated by the absence of insulin in untreated Type I Diabetics. What happens to insulin levels in alcoholics? Hint #2 We are told that alcoholics have limited glycogen stores AND that levels of NADH increase. Increased NADH/NAD+ will DECREASE flow through Krebs Cycle, which should increase the amount of "excess" acetyl CoA that can contribute to ketone synthesis. Hint #3 It is hypothesized that ethanol may actually INCREASE hydrolysis of triacylglycerides from adipose cells, which will increase the production of acetyl CoA, and, thus ketones cannot be correct.

G6PD deficiency affects all cells in the body, but RBCs are particularly vulnerable. Which of the following is the most plausible explanation for this? Please choose from one of the following options. A. Hemoglobin carries oxygen and, thus, is less susceptible to reactive oxygen molecules. B. Other tissues have alternate pathways to generate NADPH in the cytoplasm that can be utilized by the cell. C. Other tissues have alternate pathways to generate NADPH inside mitochondria that are not altered by the deficiency. D. Hemoglobin carries oxygen and, thus, is more susceptible to reactive oxygen molecules.

More oxygen means a greater probability of reactive oxygen molecules forming! Hint #2 RBCs lack mitochondria. Hint #3 Other tissues have alternate pathways to generate NADPH inside the mitochondria is the most plausible since RBCs lack mitochondria. Note that RBCs also have no nucleus and therefore cannot generate more of enzyme in cases where the enzyme produced may be functionally impaired.

A deficiency in what other enzyme would most closely mimic the clinical manifestations of a CPTII enzyme deficiency? Please choose from one of the following options. Glucose-6-phosphatase Hormone-sensitive lipase Pyruvate carboxylase Fatty acid synthase

What does CPTII do? It transports long-chain fatty acids into the mitochondria from the cytoplasm for oxidation. Therefore, anything that affects the oxidation of fatty acids is likely to have similar manifestations to a CPTII deficiency. Hint #2 Pyruvate carboxylase and Glucose-6-phosphatase are enzymes involved in gluconeogenesis. Deficiency of these may cause hypoglycemia, but this is not the best answer choice as additional symptoms will be present. Hint #3 Fatty acid synthase is important in the synthesis of fatty acids, not oxidation. Hint #4 Presumably, if Hormone-sensitive lipase was defective and did not release fatty acids in response to decreased insulin and increased glucagon levels, it would mimic the effects of CPTII enzyme deficiency because fatty acid oxidation would be impaired. It may even cause more severe symptoms since even oxidation of medium and small chain fatty acids would be impaired.

Hormone-sensitive lipase is very sensitive to insulin levels in the body. With this in mind, what is a plausible reason for why Type I diabetics are more susceptible to ketoacidosis than Type II diabetics? Please choose from one of the following options. A. Hormone-sensitive lipase activity is more irregular in type II than type I diabetics B. Hormone-sensitive lipase is more active in type II than type I diabetics C. Hormone-sensitive lipase activity is more irregular in type I than type II diabetics D. Hormone-sensitive lipase is more active in type I than type II diabetics

What does Hormone-sensitive lipase do? How is it regulated? Hint #2 Hormone-sensitive lipase is an enzyme in adipose cells that hydrolyzes fatty acids from stored triacylglycerides. It is activated hormonally by a decrease in insulin and increase in glucagon. Hint #3 Because Type I diabetics do not produce any insulin, their hormone sensitive lipase is always "ON" - this can cause an increase in fatty acid oxidation and, thus, rising ketone levels. Type II Diabetics have at least some level of insulin production, which is probably why their hormone-sensitive lipase is kept in check.

Administration of glucagon to a symptomatic individual with a CPTII deficiency (can't oxidize fatty acids) is predicted to: Please choose from one of the following options. Worsen a patient's symptoms Improve a patient's symptoms Not change a patient's symptoms The effect on patient's symptoms cannot be predicted

What does glucagon do? Recall that it stimulates gluconeogenesis and fatty acid oxidation. Hint #2 In CPTII deficiency, there is a build-up of unused fatty acids, which presumably causes the muscle weakness and pain in some individuals. Moreover, the inability to effectively oxidize fatty acids will impair gluconeogenesis, which is reliant on the ATP produce in fatty acid oxidation. Hint #3 Increased levels of glucagon will stimulate further fatty acid release from adipose cells and maintain the overall "catabolic" state of the body. Because individuals with CPTII deficiency cannot effectively oxidize fatty acids, increased release of fatty acids will cause an increased build-up of fatty acids and continued impairment of gluconeogenesis, which will likely worsen the patient's symptoms.

An inhibitor of Complex II in the ETC would cause all of the following EXCEPT: Please choose from one of the following options. A. An unchanged proton concentration in the inter-membrane space B. Potential damage to cells from reactive oxygen species C. A build-up of FADH2 D. Impaired oxidative phosphorylation

What occurs at Complex II? Hint #2 FADH2 donates its electrons at Complex II. Notably, it is also downstream from Complex I (where NADH donates its electrons). Thus, it will impair electron flow from both NADH- and FADH2-donated electrons. Therefore, "Impaired oxidative phosphorylation" is a correct choice since ATP production is dependent on electron flow through the complexes. Hint #3 This is tricky. Although Complex II doesn't directly contribute to the proton gradient, it is decreasing electron flow to the downstream complexes (III and IV) that DO contribute to the protein gradient. Therefore, the proton gradient will decrease (it has to for ATP production to decrease as well). Note that a build-up in oxygen that normally is reduced by electrons in Complex IV will cause an increase in reactive oxygen species. Also note FADH2 can't donate its electrons to Complex II anymore and therefore will not be re-oxidized to FAD.

What is the function of adenylyl cyclase? Please choose from one of the following options. Convert ATP to ADP Convert ADP to cAMP Convert ATP to cAMP Convert ADP to AMP

cAMP stands for cyclic adenosine monophosphate. Hint #2 ADP is not present in high concentrations in our cell. Hint #3 Adenylyl cyclase, also known as adenylate cyclase, is an enzyme that converts adenosine triphosphate to cyclic adenosine monophosphate.