AAMC - missed questions biology

B. Increase the intraneuronal NE concentration Inhibiting MAO would prevent the breakdown of NE within the neuron, leaving a greater amount of NE to be stored in vesicles, resulting in an increase in the intraneuronal NE concentration ****whenever evaluating what happens when something is inhibited, first identify how it should work then flip every downstream affect

A compound that inhibits monoamine oxidase (MAO) should have which of the following effects on NE concentration? A. Increase the extraneuronal NE concentration B. Increase the intraneuronal NE concentration C. Decrease the intraneuronal NE concentration D. Decrease the extraneuronal NE concentration

B. more than two sets of chromosomes. spindle fibers should separate chromosomes, so in their absence, the chromosomes would stay on one side and we would end up with two sets

A drug that binds to tubulin molecules of plant cells and prevents the cells from assembling spindle microtubules would most likely cause the resulting plants or plant cells to have: A. greater genetic variability than the parent plants. B. more than two sets of chromosomes. C. a stronger cell wall because of excess tubulin. D. independent movement because of excess tubulin.

C. A decrease, because the decreased rate of urine output will allow more reabsorption by the kidney The best answer is answer choice C: that low blood pressure decreases the glomerular filtration rate, allowing more time for reabsorption and decreasing the amount of substance A in the urine.

A lower-than-normal blood pressure will cause which of the following effects on the rate of plasma clearance of Substance A? A. An increase, because the concentration of Substance A in the urine will increase B. An increase, because the ADH levels will be very low C. A decrease, because the decreased rate of urine output will allow more reabsorption by the kidney D. A decrease, because ADH levels will be very high

C. a reduction in FSH concentration. **sertoli cells = spermatogenesis

A male taking excess testosterone may become infertile because of reduced spermatogenesis. According to Figure 2, this could result directly from: A. an increase in inhibin concentration. B. a reduction in inhibin concentration. C. a reduction in FSH concentration. D. a reduction in LH concentration.

C. approximately the same. The passage only states that post-menopausal women show the accelerated bone loss and the acceleration slows after eight to ten years. From this it seems likely that the acceleration has something to do with withdrawal of estrogen. Administration of estrogen to men would then have no therapeutic effect on bone loss, since they should not be suffering from withdrawal of the hormone, having never had high levels. Men given estrogen should have the same chance of developing osteoporosis as a control population,

A man is treated with low doses of an estrogen analogue to destroy an estrogen-responsive adrenal tumor. Compared to an age-matched control (no estrogen treatment), this patient's chances of developing osteoporosis will most likely be: A. increased. B. decreased. C. approximately the same. D. approximately the same, but the disease will appear at an earlier age.

B.a carboxyl group. fatty acids contain a carboxylic acid head group and a hydrocarbon tail.

A fatty acid is comprised of a long hydrocarbon tail and a head consisting of: A.a hydroxyl group. B.a carboxyl group. C.a phosphate group. D.an amino group

B. III only pathogenic disease: contagious because the pathogen itself is able to spread from one host to another nonpathogenic diseases will not be contagious * The passage states that suspicion of a microbial agent was established before the virion was found. Option I can not be correct because it refers to particles found in the blood. Viruses can not be cultured in media so option II can not be correct. The fact that the disease was infectious was sufficient to put forth that hypothesis.

A microbe pathogen was hypothesized as the causative agent of the disease described in the passage because: I. suspicious objects were found in blood samples from ill patients. II. in vitro cultivation of the probable pathogen was difficult. III. the disease was infectious. A. II only B. III only C. I and II only D. I and III only

C.The cell can bind to endothelium.

A neutrophil has point mutations in the genes coding for the alpha and the beta subunits of the adhesion receptor. However, this cell can still migrate through endothelium. Which of the following conclusions about the effect of this mutation can be drawn? A.The cell cannot release toxic products such as prostaglandins. B.The cell has only functional beta subunits. C.The cell can bind to endothelium. D.The cell has a defective cell membrane

B. breakdown of the body's own structural proteins to provide energy. In starvation, the body uses up its stores of carbohydrate and lipids, and then begins to break down body proteins for metabolic energy. A byproduct of the metabolism of the amino acids from protein is nitrogen. ** A byproduct of the metabolism of the amino acids from protein is nitrogen** the nitrogen in the urine comes from breakdown of the body's proteins. (Incomplete reabsorption of nitrogenous products is the goal of the kidney, not necessarily a pathological process.- why D is wrong)

A resident of a famine area who appears undernourished and extremely emaciated has eaten only starches for the past 3 months. A urine analysis shows that a large amount of nitrogen is being excreted. This is most likely evidence of: A. an abnormally high rate of glycogen breakdown in the liver. B. breakdown of the body's own structural proteins to provide energy. C. utilization of the last remaining fat reserves to provide energy. D. incomplete reabsorption of nitrogenous products due to kidney failure.

B. Smooth Enhanced activity of smooth muscles in blood vessels would cause vasoconstriction increased vasoconstriction is a major cause of hypertension

According to Hypothesis A, enhanced activity of which of the following basic muscle types would be most likely to cause hypertension? A. Striated B. Smooth C. Cardiac D. Multinucleate

B. respiration. in mammals, skin is not responsible for exchanging oxygen and carbon dioxide with the environment** so B is wrong

All of the following are functions of mammalian skin EXCEPT: A. sensation. B. respiration. C. protection from disease. D. protection against internal injury.

A. G1 the cell cycle stage in which a stable, differentiated, nondividing cell will most likely be found: A diploid, nondividing cell is most likely in G0 or G1, in which the cell remains metabolically active** (but not replicating DNA -S, or segregating its duplicated chromosomes and dividing -M)

A stable, differentiated cell that will NOT divide again during its lifetime would most likely be found in which of the following stages of the cell cycle? A. G1 B. G2 C. M D. S

D. Ki67 Platelets are cell fragments without nuclei and therefore would not be expected to contain a protein like Ki67 that is detected exclusively in the nuclei of proliferating whole cells. Because this option presents a situation that is unlikely to be true, it is the correct answer to the question.

According to the passage, platelets are LEAST likely to contain: A. transmembrane serotonin transporters. B. ribosomes. C. serotonin. D. Ki67

C. 20,000 The passage states that 20 percent of T lymphocytes are activated by superantigens, while one in 100,000 is activated by conventional antigens. Taking 20 percent of 100,000 yields 20,000 MATH ERROR***

According to the passage, superantigens increase the number of activated T cells over activation levels observed with conventional antigens by a factor of: (Because this unique type of binding activates approximately 20% of the T lymphocytes, as opposed to 1 in 100,000 T cells activated by conventional antigenic stimulation, superantigens are considered nonspecific stimulators) A. 20. B. 5,000. C. 20,000. D. 100,000.

D. II and III only You would expect to find ice in the extracellular fluid - i.e., blood plasma (II) and lymph (III) in the frozen body of a freeze-tolerant frog. However, ice in the cytoplasm (III) would be lethal because ice crystal formation within the cells disrupts structural organization. Thus, answer choice D is the best answer

According to the passage, which of the following parts in the frozen body of a freeze-tolerant frog would contain ice? I. Cytoplasm II. Blood plasma III. Lymph A. II only B. III only C. I and II only D. II and III only

D. sperm motility. sperm motility requires large amounts of ATP as evidenced by the high concentration of mitochondria in the sperm midpiece. Therefore, of the choices given, reduced sperm motility

Accumulation of DDT in the testes may cause reduced fertility in males because the uncoupling of oxidative metabolism from ATP production may reduce: A. glucose concentration of semen. B. testosterone concentration of semen. C. blood circulation in the testes. D. sperm motility.

C.CO2 in the tissues. inhibition of carbonic anhydrase would slow down the production of carbonic acid from water and carbon dioxide to the uncatalyzed reaction rate. Loss of this part of the reaction would allow for the buildup of carbon dioxide in the tissues because the normal method of carbon dioxide removal is inhibited.

Administration of a carbonic anhydrase inhibitor to the RBCs would most likely cause an increase in the concentration of: A.HCO3- in the RBCs. B.H2CO3 in the RBCs. C.CO2 in the tissues. D.H2O in the tissues.

D. Decreased volume and decreased pressure An aldosterone deficiency would result in lower Na+ reabsorption into the bloodstream. Because H2O passively follows Na+ during reabsorption in the kidney, less Na+ reabsorption would result in less H2O reabsorption into the bloodstream. This would result in decreased blood volume. Blood volume would also be affected by lower blood Na+ levels because there would be less of this ion to osmotically hold water in the extracellular fluid. Decreased blood volume would result in decreased blood pressure as well

Aldosterone stimulates Na+ reabsorption by the kidneys. What changes in blood volume and pressure would be expected as a result of aldosterone deficiency? A. Increased volume and increased pressure B. Increased volume and decreased pressure C. Decreased volume and increased pressure D. Decreased volume and decreased pressure

A.eukarya. the passage states that BFA disrupts the Golgi apparatus, and of the choices, only eukaryotes have a Golgi apparatus***

Based on the mode of action described for BFA in the passage, the drug would be most effective against: (Brefeldin A (BFA), a lactone compound isolated from fungi, has been shown to inhibit Arf1-driven vesicle formation, resulting in reversible disruption of the Golgi apparatus and tumor remission in vitro.) A.eukarya. B.viruses. C.bacteria. D.archaea.

A. low concentrations of DNA in the macronucleus. For the extra S phase to be advantageous to the organism it must, in some way, be responsive to low DNA levels,

An extra S phase occurs during amitotic division in a small macronucleus to minimize fluctuations in DNA content. This is most likely triggered by the presence of: A. low concentrations of DNA in the macronucleus. B. centromeres in the macronucleus. C. high concentrations of DNA in the micronucleus. D. mitotic enzymes in the micronucleus.

B. the peritoneal cavity. Chris*

An ulcer that penetrated the wall of the intestine would allow the contents of the gastrointestinal tract to enter: A. the perineum. B. the peritoneal cavity. C. the pleural cavity. D. the lumen of the intestine.

B. Increase the number of visual fields counted per petri dish. random sampling always increases in accuracy if the number of samples analyzed is higher** **look for increase in sample size

As one step in the estimation of the efficiency of neuronal induction, scientists calculated the average number of induced cells present in 30 randomly selected 20× visual fields. Which change to this particular aspect of the experimental protocol would increase the accuracy of the estimates of efficiency? A. Increase the magnification of the oculars used to define the field of view. B. Increase the number of visual fields counted per petri dish. C. Select visual fields from the central portion of the petri dish where cell density is highest. D. Use the presence of green fluorescence to identify cells appropriate for quantification.

C. At any concentration greater than 1 µM **SILLY only time the + end is growing is after 1µM so has to be after than that

At what concentration of free actin will the + end of the microfilament grow faster than the - end? A. Exactly at 1 µM B. Only between 1 µM and 4 µM C. At any concentration greater than 1 µM D. At any concentration

A.Blood glucose levels wild-type mice produce more proinsulin in response to a glucose injection and process more proinsulin to insulin than do Cdkal1-/- mice. Insulin stimulates cellular uptake of glucose, and higher cellular uptake of glucose would result in lower blood glucose levels.

Based on the passage, the level of which of the following is most likely lower in nonfasting wild-type mice than in nonfasting Cdkal1-/- mice? A.Blood glucose levels B.Cellular glucose uptake C.Liver glycogen synthesis D.Cellular protein synthesis

B. The - ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. Within a sarcomere, the microfilament length remains stable. Because one end of the microfilament is anchored in the Z line, actin monomers are prevented from being added to or subtracted from that end. This rules out the possibility of treadmilling. Therefore, to retain a stable length, both ends of the microfilament must be capped.

Below is a diagram of a muscle sarcomere. Based on the passage, which statement best explains why the microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction? A. The - ends of the microfilaments are capped by Z lines, and the actin subunit concentration is kept above 1 µM in muscle cells. B. The - ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. C. The actin subunit concentration is kept above 4 µM in muscle cells. D. The - ends polymerize and the + ends depolymerize at the same rate.

A. Net fluid flow in the direction of interstitial spaces will increase. blood clot leads to changes in hydrostatic pressure by preventing blood flow **build up of hydrostatic pressure in the capillaries = blood volume leaks into interstitial fluid

Capillaries in the kidney and elsewhere in the body maintain fluid homeostasis by balancing hydrostatic and osmotic pressures. Which of the following is the initial effect of a blood clot forming on the venous side of a capillary bed? A. Net fluid flow in the direction of interstitial spaces will increase. B. Net fluid flow in the direction of interstitial spaces will decrease. C. Capillary osmotic pressure will increase. D. Capillary osmotic pressure will decrease.

B. Inhibition of leukocyte phagocytosis of uric acid crystals Colchicine affects microtubule organization and the only answer choice that depends on microtubules is: B - phagocytosis phagocytosis requires endosome transport to the lysosome, so it relies on microtubules

Colchicine most likely relieves gout symptoms through what mechanism? A. Prevention of uric acid diffusion through cell membranes B. Inhibition of leukocyte phagocytosis of uric acid crystals C. Inhibition of uric acid crystal formation D. Maintenance of the pH optimum for PRPP synthetase

A. more negative IPP and inspiration. negative pressure is what allows for inspiriation and is the more active process where the diaphragm contracts constriction = inspiration

Contraction of the diaphragm results in a: A. more negative IPP and inspiration. B. more negative IPP and expiration. C. more positive IPP and inspiration. D. more positive IPP and expiration.

A. individuals with UV-reflective dewlaps produced more offspring than did individuals without them. To evolve by natural selection and become a general characteristic of the species, the genes that cause dewlaps to reflect UV light must become a significant portion of the gene pool, which will most likely occur if individuals with UV-reflective dewlaps produce more offspring than do individuals without them. ** passing on the genes that cause the advantageous phenotype**

Dewlaps that reflect UV light would evolve by natural selection only if: A. individuals with UV-reflective dewlaps produced more offspring than did individuals without them. B. individuals with UV-reflective dewlaps were better able to communicate than individuals without them. C. individuals with UV-reflective dewlaps were less subject to predation than individuals without them. D. individuals with UV-reflective dewlaps mated more frequently than did individuals without them.

D. Use an alternate antibiotic. A moderate increase in the temperature should increase the translation efficiency. Thus, use of an alternate antibiotic is the only reasonable choice of those presented. Thus, D is the correct response.

Chloramphenicol did NOT inhibit translation in E. coli cells containing the cat6xbs expression plasmid. What experimental parameter could be changed in order to affect translation inhibition? A. Increase the chloramphenicol concentration. B. Increase the chloramphenicol incubation time. C. Alter the incubation temperature by a few degrees. D. Use an alternate antibiotic.

D. could be on a sex chromosome or on an autosome The fact that the pigment is expressed in the dewlap, a structure found only in males, is not sufficient to eliminate any chromosome as the location of this gene. ***

If Anolis lizards have X-Y chromosomal sex determination, the locus of a gene for the UV reflectance pigment: (Male lizards have a dewlap, a large fold of skin under the throat that they can fan out like a flag.) A. must be on the X chromosome. B. must be on the Y chromosome. C. must be on an autosome. D. could be on a sex chromosome or on an autosome

A. Extinction ***only females so extinction

If all genotypes are equally fit and if there are no genetic modifiers of the sex ratio trait, what will be the ultimate fate of a population in which 50% of the X chromosomes are currently Xi and 50% are Xs? (passage states that if none of the Xi genotypes are selected against, then the Xi chromosome will increase to 100%. - An XiY male expresses the sex ratio trait: he sires only daughters. ) A. Extinction B. Stable population size, with a predominance of females C. Stable population size, with all individuals producing a 50:50 sex ratio D. Stable population size, with some individuals producing an excess of females and some producing an excess of males

C. Hypothesis B, because the kidneys were responding to decreased glomerular blood pressure

If restriction of blood flow to the kidneys (by placing clamps on the renal arteries) resulted in an immediate but small increase in blood pressure, followed by the gradual development of severe hypertension, which hypothesis would these results best support? A. Hypothesis A, because the clamps increased the vascular resistance to blood flow B. Hypothesis A, because the clamps caused the kidneys to receive less blood C. Hypothesis B, because the kidneys were responding to decreased glomerular blood pressure D. Hypothesis B, because the volume of body fluids was probably decreasing

D. the surface of the proteins encoded by the genes for the disease. Antigens are carried on the surface of cells, not on the chromosomes, DNA segments, or RNA.**

If the genetic and autoimmune theories of inflammatory bowel disease are true, then the gastrointestinal antigen being targeted by the immune system is probably on: A. the chromosomes carrying the genes for the disease. B. part of the DNA segments constituting the genes for the disease. C. stretches of the mRNA's coded for by the genes for the disease. D. the surface of the proteins encoded by the genes for the disease.

B. EMS to E. when considering differentiation, think about the step where cell has lost the ability to be any thigb other than the set of cell types being considered (only considering gut differentiation) If the zygote contains all unique cell contents that are necessary for gut differentiation, segregation of these substances during cell division would occur in the sequence of zygote to P1 to EMS to E. This information comes directly from the flow chart in Figure 1, which shows that gut cells are derived from the following source: zygote → P1 → EMS → E.

If the zygote contains unique cell contents that are necessary for gut differentiation, segregation of these substances during cell divisions would occur in the sequence of zygote to P1 to: A. AB. B. EMS to E. C. P2 to EMS to E. D. both P2 and EMS

A. glomerular filtration rate. increased blood pressure will affect the glomerular filtration rate (Water resorption and concentrating ability are the same, so answer choices C and D are essentially the same. so can rule them out)

In Experiment 2, the increased blood pressure resulting from the higher-than-normal concentration of ADH most likely affected the urinary output of Substance A by increasing the: A. glomerular filtration rate. B. Tm of solutes. C. water reabsorption from the tubules. D. concentrating ability of the loop of Henle.

B. Chromosome The question asks the examinee to identify the structure that when damaged would most likely lead to cancer or result in a mutation. Mutations are heritable changes in the sequence of the nucleic acid component of chromosomes (B), and mutations that lead to unregulated cellular growth can lead to cancer

DDT would most likely initiate cancer or cause a mutation if which of the following structures is damaged? A. Nuclear envelope B. Chromosome C. Ribosome D. Histone

D. No, because the cells obtained lack the correct neurotransmitter phenotype. in order to fix parkinson's the neurons would need to produce dopamine** these do not Although the cells obtained under the experimental conditions are similar to CNS neurons and may therefore be similar to neurons found in the brain, the neurons that are generated produce either glutamate or GABA. Since Parkinsonism is associated with deficiency of dopamine-producing neurons, neither of these 2 types of cells would alleviate the symptoms of this disorder.

Does the experimental approach described in the passage yield cells that could be used in an animal model of Parkinson disease to replace dopamine-deficient neurons in the brain? A. Yes, because the cells obtained have the functional characteristics of nerve cells. B. Yes, because the cells obtained are similar to cells in the central nervous system. C. No, because the cells obtained may contain tumorigenic pluripotent cells. D. No, because the cells obtained lack the correct neurotransmitter phenotype.

D. The voltage-gated K+ channels are open, and the voltage-gated Na+ channels are closed. repolarization = the movement of the membrane potential from a less polarized to a more polarized state due to an influx of K+ and leads to hyperpolarization

During the repolarization phase of an action potential in a neuron, which of the following is generally true of the voltage-gated channels that cause that action potential? A. The voltage-gated K+ channels are closed, and the voltage-gated Na+channels are closed. B. The voltage-gated K+ channels are open, and the voltage-gated Na+ channels are open. C. The voltage-gated K+ channels are closed, and the voltage-gated Na+channels are open. D. The voltage-gated K+ channels are open, and the voltage-gated Na+ channels are closed.

C. increased at least 10 times by the presence of the endothelium Relaxation occurs in the ring with endothelium at 10-7 M ACH but in the other ring does not occur even with 10-6 M ACH; these concentrations differ by a factor of 10. Both rings show the same response to norepinephrine. There is no response to ACH at 10-8 M with or without endothelium. Thus, answer choice C is the best answer.

From the data in Figure 1, one can conclude that the sensitivity of aortic smooth muscle to acetylcholine is: A. decreased by the presence of norepinephrine. B. increased by the presence of norepinephrine. C. increased at least 10 times by the presence of the endothelium. D. greatest at 10-8 M, with or without endothelium.

B. III only The heart and blood vessels both differentiate from the mesoderm.***

From which germ layer(s) do the tissues of the heart and blood vessels differentiate? I. Ectoderm II. Mesoderm III. Endoderm A. II only B. III only C. I and II only D. I and III only

C. 40% All the males in the next generation will acquire a Y chromosome from their male parent, so the contribution of the male can be ignored in solving this problem. All the XiXi females will have XiY sons, so 15% of the XiY flies in the next generation will come from this type of female. Half of the sons the XiXs females produce will be XiY. Since XiXs females make up 50% of the population, 25% of the males in the next generation will come from this type of female. None of the sons of XsXs females will be XiY. The total number of male flies that are XiY is 40%

In a laboratory population of Drosophila, all the males are XsY. Among the females, 15% are XiXi, 50% are XiXs, and 35% are XsXs. Assuming random mating, what proportion of male flies in the next generation will be XiY? A. 12% B. 30% C. 40% D 65%

B. The coordination of cell differentiation during development The destruction of mRNA prevents continuous protein production, allowing the cell to change its protein expression over time. B is the best answer because the coordination of cell differentiation during development is extremely sensitive to the timing of mRNA turnover.

If oligonucleotides such as mRNA were not degraded rapidly by intracellular agents, which of the following processes would be most affected? A. The production of tRNA in the nucleus B. The coordination of cell differentiation during development C. The diffusion of respiratory gases across the cell membrane D. The replication of DNA in the nucleus

A.In euchromatin GAPDH is a housekeeping gene and is expressed continuously. A gene that is always turned on must be accessible to transcription factors. Only euchromatin is in a loose conformation and readily accessible for transcription.*** SILLY--> even though GAPs turn things off, is something is continuously expressed it has to be in EUCHROMATIN

If the GAPDH gene is continuously expressed, where is it most likely found? A.In euchromatin B.In a telomere C.In heterochromatin D.In a centromere

C. S phase Generally, if a site is broken that would be the step we get stuck in**

If the anti-inflammatory drug in Treatment 1 interfered with DNA replication, in which phase of the cell cycle would cells tend to be arrested? A. G0 phase B. G1 phase C. S phase D. G2 phase

A. fertilization and birth only. female life cycle: fertilization, birth, puberty, menopause All of the mitotic divisions that form primary oocytes occur prior to birth so A

In human females, mitotic divisions of oogonia that lead to formation of presumptive egg cells (primary oocytes) occur between: A. fertilization and birth only. B. fertilization and puberty only. C. birth and puberty only. D. puberty and menopause only.

B. The splitting of centromeres One of the key differences between mitosis and meiosis occurs during their respective anaphases. During anaphase of mitosis, sister chromatids are pulled apart at the centromeres, each becoming an independent chromosome in the two diploid daughter cells. During anaphase I of meiosis I, homologous pairs of chromosomes are separated into the two daughter cells. However, each chromosome still consists of two sister chromatids joined to each other at the centromere. ****It is not until anaphase II of meiosis II that the centromere is split and the sister chromatids separate

In mammals, which of the following events occurs during mitosis but does NOT occur during meiosis I? A. Synapsis B. The splitting of centromeres C. The pairing of homologous chromosomes D. The breaking down of the nuclear membrane

C. ulcers could be produced in healthy organisms by infecting them with H. pylori. One of these criteria is the demonstration that the putative infectious agent is capable of causing the disease or condition in an organism that was healthy prior to exposure to the agent. C correctly identifies this criterion by stating that proof that H. pylori can cause stomach ulcers depended on demonstrating that ulcers can be produced in healthy organisms by infecting them with H. pylori **ESTABLISH CAUSALITY

It is now generally accepted that H. pylori can cause ulcers. Proof of this most likely depended on the demonstration that: A. people with stomach ulcers have antibodies to H. pylori. B. healthy individuals have antibodies to H. pylori. C. ulcers could be produced in healthy organisms by infecting them with H. pylori. D. the organism can be passed from mother to fetus during pregnancy.

B. 25% The macronuclei do not participate in mating so only the genotypes of the micronuclei need to be considered. If we call the recessive gene r and its dominant allele R, then a cross between two heterozygote strains Rr will produce the genotypes RR:Rr:rr in a ration of 1:2:1.

In a mating of two Tetrahymena strains that are homozygous in their macronuclei and heterozygous in their micronuclei for a recessive gene, what percentage of the F1 generation will express the recessive phenotype? A. 0% B. 25% C. 50% D. 100%

A.a 1-electron carrier. cytochrome c is a heme protein that only cycles between a ferrous and ferric state during oxidative phosphorylation. Therefore, only single electron transfers are possible.

In oxidative phosphorylation, cytochrome c acts as: A.a 1-electron carrier. B.a 2-electron carrier. C.a 3-electron carrier. D.a 4-electron carrier.

A. The nucleus Uridine is incorporated into RNA in the nucleus where transcription of DNA into RNA takes place. RNA is manufactured in the nucleus from a DNA template. Therefore, the correct answer is answer choice A.

In which organelle of a eukaryotic cell is the pyrimidine uracil, as part of uridine triphosphate (UTP), incorporated into nucleic acid? A. The nucleus B. The Golgi bodies C. The ribosomes D. The endoplasmic reticulum

D. negative pressure pumping action. Inflation of lungs in mammals is accomplished by negative pressure pumping action.

Inflation of the lungs in mammals is accomplished by: A. diffusion of gases. B. active transport of gases. C. positive pressure pumping action. D. negative pressure pumping action.

C. Periodic menstruation will resume. Estrogen and progesterone are actively secreted by the ovaries of pre-menopausal women and act to maintain the uterine cycle. With advancing age the ovary becomes less responsive to pituitary gonadotropins and cyclical changes in the endometrium of the uterus disappear. The menstrual cycle can be re-established by administration of estrogen and progesterone in a regimen that approximates the rise and fall of hormone levels in pre-menopausal women.

Postmenopausal women receiving estrogen and progesterone therapy will most likely experience which of the following side effects? A. Breast tissue will atrophy. B. Vaginal tissue will dry out. C. Periodic menstruation will resume. D. Lactation will be induced

A. inadequate blood volume for effective filtration. This question asks the examinee to identify the cause of kidney failure during times of extreme dehydration As the passage states, severe dehydration greatly reduces the volume of filtrate moving through the nephrons of the kidney. If fluid volume is too drastically reduced, the kidney will be unable to effectively do its job of filtering and maintaining homeostasis within bodily fluids.

Kidney failure during severe dehydration is most likely due to: (During dehydration, the kidneys may reduce their urinary output from the normal level of 1.0-1.5 L H2O/day to as little as 0.5 L H2O/day, and renal salt excretion may decline to near zero.) A. inadequate blood volume for effective filtration. B. inability to produce sufficient urine. C. buildup of salts in the distal tubules. D. increased body temperature.

A. the ability to produce ATP via ATP synthase. Both types of cells possess a membrane-embedded electron transport chain capable of generating a H+ gradient, which drives synthesis of ATP via ATP synthase.

Most bacterial cells and human cells are alike in: A. the ability to produce ATP via ATP synthase. B. the chemical composition of their ribosomes. C. their enclosure within cell walls. D. the shape of the self-replicating structures that carry their DNA.

D. it suppresses cells specific to the body's own antigens The immune system is designed to attack foreign material in the body. It avoids attacking tissues of its own body because it suppresses cells that are specific to its own body's antigens (surface molecules that would otherwise initiate an immune response).

Normally the immune system avoids attacking the tissues of its own body because: A. a special intracellular process recognizes only foreign antigens. B. the body does not make any antigens that the immune system could recognize. C. it changes its antibodies to be specific only to foreign antigens. D. it suppresses cells specific to the body's own antigens.

A.high, because Equation 1 will proceed to the right. Equation 1 leads to the conclusion that increasing the concentration of carbon dioxide will result in higher levels of H+ (lower pH) based on LeChâtlier's Principle. The lower pH will increase the dissociation of oxygen.

O2 dissociates more readily from Hb in an acidic environment. This dissociation will therefore occur most readily when the PCO2 is: A.high, because Equation 1 will proceed to the right. B.high, because Equation 1 will proceed to the left. C.low, because Equation 1 will proceed to the right. D.low, because Equation 1 will proceed to the left.

D. raising the environmental temperature. Water is lost through the skin primarily as a means to keep the body at normal temperatures. Therefore, raising the environmental temperature would cause a person to perspire, releasing water to the environment where it can evaporate and cool down the body.

The most effective method for producing an increase in the total amount of water lost through the skin during a certain period would be: A. inhibiting kidney function. B. decreasing salt consumption. C. increasing water consumption. D. raising the environmental temperature.

C. regulated in a similar manner as the target gene. To provide effective therapy, this antisense gene would need to be regulated in a manner similar to the manner in which the target gene is regulated so that the antisense RNA is produced at the same time that the sense mRNA is produced. *** This would ensure that the antisense RNA is available to bind with the sense mRNA, thereby preventing its subsequent translation.

To be an effective therapy, an antisense gene that is incorporated into a genome that contains the target gene must be: A. on the same chromosome as the target gene but not necessarily be physically adjacent. B. on the same chromosome as the target gene and must be physically adjacent. C. regulated in a similar manner as the target gene. D. coded on the same strand of DNA as the target gene.

C. They should have circular DNA Such a characteristic is likely to have been highly conserved over time since it involves the basic material of life.**

To support the symbiotic hypothesis presented in the passage, mitochondria should be similar to bacteria in which of the following ways? A. They should use 80S ribosomes. B. They should be incapable of binary fission. C. They should have circular DNA. D. They should be capable of anaerobic respiration.

C. 6500 mL The amount of air entering the lungs in a single breath, or tidal volume, is given as 800 mL/breath. Of that 800 mL only 650 mL reaches the alveoli per breath (800 mL of air inhaled minus 150 mL of nonalveolar respiratory volume). Therefore the net volume of air that reaches the alveoli each minute is equal to 650 mL/breath multiplied by 10 breaths/min, or 6500 mL.

What is the net volume of fresh air that enters the alveoli each minute, assuming that the breathing rate is 10 breaths/min, the tidal volume is 800 mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL? A. 65 mL B. 95 mL C. 6500 mL D. 7850 mL

C. RNA translation. When antisense mRNA molecules are used as described in the passage, the antisense molecules bind specifically to the sense mRNA, preventing the process of translation the passage describes the effect of antisense RNA binding to mRNA, and mRNA is directly involved in translation. can do A and C, but passage is specific "prevent the production of the natural gene product.""

When used as described in the passage, antisense drugs prevent: A. DNA replication. B. RNA transcription. C. RNA translation. D. cell replication

C. II and IV only Sertoli cells support and nourish the spermatocytes and promote the process of spermatogenesis. Spermatogenesis would not occur without Sertoli cells. The two hormones that directly stimulate Sertoli cells are FSH and testosterone.

Which of the following hormones is(are) directly required for spermatogenesis? I. Luteinizing hormone (LH) II. Follicle-stimulating hormone (FSH) III. Inhibin IV. Testosterone A. IV only B. I and IV only C. II and IV only D. I, II, and III only

D. A tissue culture Viruses take advantage of the elaborate intracellular mechanisms of the host cell using them to make more virus particles. To do this they need intact host cells. Non-cellular media such as the nutrient broths and suspensions in choices A, B and C will not support culture of viruses. They can be grown in tissue culture, so choice D is the correct answer.

Which of the following media would most likely be used to grow virions in the laboratory? A. A suspension of ribosomes and ATP B. A suspension of human DNA C. A nutrient broth D. A tissue culture

B. gain, rather than lose, heat by radiation. If people lack sweat glands, they are unable to make sweat nor to capitalize upon the evaporative cooling of sweat (A - wrong). These individuals are forced to rely nearly solely on vasodilation (radiation) for responding to elevated external temperatures. That individuals without sweat glands are likely to suffer heat stroke in the tropics, indicates that radiation alone is ineffective for cooling under these conditions.

People who are born without sweat glands are likely to die of heat stroke in the tropics. This indicates that, under tropical conditions, the human body may: A. gain, rather than lose, heat by evaporation. B. gain, rather than lose, heat by radiation. C. need to use different mechanisms than in temperate zones to maintain body temperature. D. be better able to regulate body temperature than under temperate conditions.

D. No, because blockage of phenylalanine hydroxylase gene expression will not remedy the original disorder The question indicates that phenylketonuria (PKU) is a genetic disorder caused by a mutation in the gene for an enzyme, which results in the elimination of enzymatic activity. The question asks the examinee to consider whether an antisense drug could help individuals who have PKU. An antisense drug works to prevent the expression of undesirable genes but does nothing to remedy the problem of a gene that produces an ineffective product. The only way to cure PKU would be to add a gene or gene product that could lead to the production of effective enzymes to replace the ones that do not function correctly.**** ( A and B are incorrect because preventing expression of an inactive protein is not a cure. C is incorrect because the stability of the mRNA is irrelevant to the situation)

Phenylketonuria is a genetic disorder caused by a mutation in the gene for the enzyme phenylalanine hydroxylase, which eliminates its enzymatic activity. Could an antisense drug help individuals with this disorder? A. Yes, if it binds to the mRNA of the phenylalanine hydroxylase gene and prevents its translation B. Yes, if it is incorporated into the chromosome and prevents the expression of the phenylalanine hydroxylase gene C. No, because mRNA does not persist in the cytoplasm of the cell D. No, because blockage of phenylalanine hydroxylase gene expression will not remedy the original disorder

B. Maintain sufficient oxygenation of cells According to the item, Sarah noticed that her skin blood vessels were usually constricted to conserve body heat in the cold environment of the Colorado mountains where she went skiing. Occasionally, however, her vessels would dilate for short periods of time to enable a sufficient supply of blood (and oxygen) to her cells. Due to the physical exertion of skiing, her cells had an increased need for oxygen.

Sarah noted that her skin blood vessels were usually constricted to conserve body heat in the cold environment of the mountains. However, her skin blood vessels would occasionally dilate for short periods of time. What would be the most probable physiological purpose for this periodic vasodilation? A. Maintain normal skin tone B. Maintain sufficient oxygenation of cells C. Reduce excessive blood pressure D. Maintain normal muscle tone

C. meiosis. The macronucleus does not go through the process of meiosis. Genes involved in meiosis are therefore superfluous in this genome. The functions of transcription (choice A), translation (choice B) and ribosome production (choice D) must be coded for by macronuclear genes because they are necessary for it to direct protein synthesis.

Some of the DNA sequences that are eliminated during macronuclear differentiation (Figure 1, Step 6) may be sequences involved in: A. transcription. B. translation. C. meiosis. D. ribosome production.

C. parasympathetic motor fibers. The autonomic nerve fibers that innervate the heart to slow it are the parasympathetic motor fibers of the vagus nerve

The autonomic nerve fibers that directly innervate the heart to cause cardiac slowing are: A. sympathetic motor fibers. B. sympathetic sensory fibers. C. parasympathetic motor fibers. D. parasympathetic sensory fibers

C. Active transport along cytoskeletal filaments B and D wrong b/c E. coli cells and, more generally, prokaryotes do not undergo mitosis AND E. coli cells do not have endoplasmic reticulum or vesicles. C allows for faster transport than diffusion

The bglF transcript is known to have a short half-life within the cytosol. What mechanism is most likely responsible for transport of this transcript to the cytoplasmic membrane once it is synthesized? (in E.coli) A. Diffusion across the cytoplasm B. Transport via attachment to the mitotic spindle C. Active transport along cytoskeletal filaments D. Transport from the endoplasmic reticulum in vesicles

D. spermatozoon. The mature ovum is the female gamete that has completed meiosis and contains the haploid number of maternally derived chromosomes. This makes it most analogous to spermatozoa, the mature male gametes that contain the haploid number of paternally derived chromosomes.

The cell type in the male reproductive system that is most analogous to the female ovum is the: A. spermatogonium. B. primary spermatocyte. C. spermatid. D. spermatozoon.

B. decrease, because movement of K+ into the mitochondrial compartments will disrupt proton movement into the intermembrane space ***DOWNSTREAM AFFECTS FIRST***

The chemical valinomycin inserts into membranes and causes the movement of K+ into the mitochondria. Based on Figure 1, if mitochondria are treated with valinomycin, the rate of ATP synthesis in the mitochondria will most likely: A. decrease, because the K+ will compete with protons at the active site on ATP synthetase. B. decrease, because movement of K+ into the mitochondrial compartments will disrupt proton movement into the intermembrane space. C. increase, because the net positive charge in the mitochondria will cause increased movement of protons into the intermembrane space. D. increase, because the additional positive charge will further activate ATP synthetase

A.DNA → mRNA the removal of introns and splicing of the remaining pieces (exons) is a step in the production of mature mRNA from the transcript produced from DNA (pre-mRNA)***

The cutting of introns and the splicing associated with the expression of the Factor VIII gene occur during which of the following steps in the protein-synthesis process? A.DNA → mRNA B.DNA → tRNA C.mRNA → tRNA D.tRNA → protein

A. Suppression of insulin secretion during freezing episodes Hyperglycemia normally elicits insulin secretion. Support for the observation that major changes in the normal glucose regulatory mechanisms occur would, therefore, be supported by the observation that insulin secretion was suppressed during hibernation -what normally prevents high glucose levels is insulin, so could be caused by lack of insulin

The extreme hyperglycemia of these animals suggests that major changes in the normal glucose regulatory mechanisms occur during freezing. Which of the following observations would support this hypothesis? A. Suppression of insulin secretion during freezing episodes B. Suppression of glucagon secretion during freezing episodes C. Slowing of glycogen catabolism in the liver during freezing episodes D. Increased sensitivity of all pancreatic endocrine responses during freezing episodes

B. one species in several different environments Adaptive radiation involves the divergence of one species into multiple species over time, which can occur when subgroups of the original species are separated or isolated in different environments so that these subgroups evolve independently of one another.

The finches observed by Darwin on the Galapagos Islands are an example of adaptive radiation. In order to set up conditions that would produce adaptive radiation, it would be necessary to place members of: A. one species in one rapidly changing environment. B. one species in several different environments. C. several very similar species in the same environment. D. several unlike species in one environment to compete for the same resources

B. marks the chromatin of the active var promoter for reexpression after mitosis According to the passage, P. falciparum cells contain the most PfSET10 when the intraerythrocyte parasites are in an actively dividing life cycle phase. To maintain cellular identity, there has to be a mechanism for marking genes that were transcriptionally active before mitosis for reactivation after mitosis. Because PfSET10 colocalizes with the active var gene, it is reasonable to hypothesize that PfSET10 methlytransferase activity is involved in this bookmarking.

The information in the passage suggests that PfSET10 has which function in var gene localization or expression? PfSET10: A. allows active and silent var genes to colocalize in the nucleus. B. marks the chromatin of the active var promoter for reexpression after mitosis. C. marks the chromatin of a silent var promoter to be expressed after mitosis. D. marks the chromatin of multiple var promoters for simultaneous expression.

A. PfEMP1 PfEMP1 is a parasite protein that is present in the plasma membrane of the RBC that the parasite inhabits. This suggests that there must be a mechanism for transporting PfEMP1 from the parasite to the RBC plasma membrane.

The information in the passage supports the prediction that P. falciparum creates unique protein trafficking structures outside the parasite itself for the trafficking of which parasite protein? A. PfEMP1 B. PfSET10 C. Histone H3 D. Hemoglobin

C. mitosis. the liver can partially regenerate after illness or damage. This regeneration is accomplished by mitosis. **

The liver is different from many other organs in that it can at least partially regenerate following illness or damage. This regeneration is accomplished primarily through: A.fission. B. meiosis. C. mitosis. D. cell growth.

A. undergo uneven division. the common feature is that both cytoplasmic division in ova and division of nuclear material in Tetrahymena are both uneven.

The macronuclei of the asexual progeny in Tetrahymena and the cytoplasm of the ova-producing cells of female vertebrates share a common feature in that both: A. undergo uneven division. B. contain uneven amounts of nuclear material. C. regulate their contents by adding or skipping an S phase. D. are apportioned at mitosis.

C. the products of fatty acid oxidation reduce skin blood flow. vitamin E makes things UNOXIDIZED (D is incorrect) antioxidants are going to interact with products of metabolism which produce free radicals, to stop the oxidative damage - will UNOXIDATE things - C is correct

The most likely explanation for the difference in skin blood flow between the fatty acid group and the fatty acid + vitamin E group in Figure 1 is that: A. vitamin E alone reduces skin blood flow more than fatty acids alone. B. vitamin E alone increases skin blood flow more than fatty acids alone. C. the products of fatty acid oxidation reduce skin blood flow. D. unoxidized fatty acids reduce skin blood flow.

C. E. coli entering the abdominal cavity from the appendix. **READ THE WHOLE PASSAGE TO SEE IF YOU CAN GET INFO** The passage states that E. coli are found within the colon, where they function in digestion and vitamin production. The appendix is continuous with the colon so that bacteria can move between these two structures; a ruptured appendix would allow E. coli into the abdominal cavity, which is not normal.****

The patient's ruptured appendix required treatment with antibiotics because he had a bacterial infection caused by: A. M. tuberculosis. B. E. coli entering the colon from the appendix. C. E. coli entering the abdominal cavity from the appendix. D. E. coli entering the appendix from the colon.

D. Protein cycloheximide is inhibiting translation and causing abnormalities, so that product is what's important (translation inhibitor = prevents production of PROTEINS at the ribosomes)

The results of Experiment 2 indicate that the signaling interaction at the two-cell stage probably most involves which class of macromolecules? (Experiment 2 Two‑cell embryos were incubated in the presence of either cycloheximide, an inhibitor of translation, or actinomycin D, an inhibitor of transcription. The AB cells were then isolated, washed to remove the inhibitors, and grown in culture. AB cells from embryos treated with cycloheximide produced only neurons and skin, whereas those from embryos treated with actinomycin D produced neurons, skin, and muscle... ***normal = neurons, skin, and muscle) A. DNA B. Messenger RNA C. Ribosomal RNA D. Protein

B.Because the antibody was generated in the mouse, repeated usage in the same patient would elicit the production of human anti-mouse antibodies. human immune system will recognize the mouse antibody as a foreign substance (antigen) and generate an immune response towards it, which may include a mild to severe allergic response.*** The immune response will generate antibodies against the mouse antibody, limiting its usefulness as a treatmen

The scientist wanted to use antibody B clinically (to treat humans), but this proposal was rejected. Which of the following is the most logical reason for the rejection? A.Because the antibody was generated in the mouse, it can never be used in humans. B.Because the antibody was generated in the mouse, repeated usage in the same patient would elicit the production of human anti-mouse antibodies. C.Because the antibody was generated in the mouse, it will not recognize human antigens. D.Because the antibody was generated in the mouse, it can only be used in vitro.

C. Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis READ GRAPH translation in interphase = increase proteolysis =decrease and decreases in mitosis

What mechanism could account for this oscillation of cyclin protein concentration? A. Replication of the cyclin gene during S phase of interphase B. Segregation of chromosomes carrying the cyclin genes during mitosis C. Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis D. Translation of cyclin mRNA in mitosis and proteolysis of cyclin protein in interphase

C. Guanine The passage states that uric acid is formed by the breakdown of purines to xanthine, a uric acid precursor. Guanine is one type of purine that is found in cells.

What nitrogenous base would promote the formation of uric acid crystals in gout? A. Cytosine B. Uracil C. Guanine D. Thymine

A. Severe neural and muscular problems due to deficiency of calcium in the plasma Removal of the parathyroid gland would lead to hypocalcemia, a condition of low blood calcium, resulting from the lack of parathyroid hormone this would cause increased neuromuscular excitability because of the change in membrane potential*** the person would eventually die from severe respiratory muscle spasms. *

What would be the result of complete removal of the parathyroid glands? A. Severe neural and muscular problems due to deficiency of calcium in the plasma B. An increase in calcitonin production to compensate for calcium deficiency in the plasma C. A drastic change in the ratio of mineral to matrix tissue in bones D. Calcification of some organs due to accumulation of calcium in the plasma

D. a variable allele distribution in the macronucleus. The amitotic division of the macronucleus will result in uneven distribution of chromosomes, hence an unpredictable genome = variable allele distribution (Given the large numbers of identical chromosomes in the macronucleus, however, repeated divisions are unlikely to result in the loss of a chromosome, so choice A is incorrect)

When an initially heterozygous macronucleus undergoes repeated binary fission, the result will be: A. the loss of macronuclear chromosomes. B. an increased rate of crossing over in the macronucleus. C. the production of a macronucleus with a genetic origin distinct from the micronucleus. D. a variable allele distribution in the macronucleus.

A.High-affinity transport (Kt) of L-alanine IF low-food conditions, high affinity transport will be important because even if the concentration of materials is low, transport can still happen quickly

Which feature of the kinetics of L-alanine transport would provide evidence that DOM is an important source of nutrients under low-food conditions? A.High-affinity transport (Kt) of L-alanine B.Low-affinity transport (Kt) of L-alanine C.High transport capacity (Jmax) of L-alanine D.Low transport capacity (Jmax) of L-alanine

C. The settling point hypothesis because, with the correct genotype, one's metabolism may allow weight to stabilize at a new level The question asks the examinee to identify which hypothesis implies that a person can deliberately alter his or her own body maintenance weight C correctly states that the settling point hypothesis allows the set point to be reset through behavior modification. (D is incorrect because it states that diet and exercise cannot reset the set point, which contradicts the premise of the settling point hypothesis.)

Which hypothesis implies that a person can deliberately alter his or her own body maintenance weight? A. The set point hypothesis because a thermostat can be reset B. The set point hypothesis because the set point can change with age C. The settling point hypothesis because, with the correct genotype, one's metabolism may allow weight to stabilize at a new level D. The settling point hypothesis because diet and exercise cannot reset the set point

B. The osmotic concentration of proteins in the dialysate fluid would increase. PROTEINS TOO BIG TO GO THROUGH MEMBRANE

Which of the following changes in flow rate or in solute concentrations would NOT occur if the blood inflow rate were increased, increasing the pressure in the dialysis chamber? A. The blood volume reaching the outflow tube per unit time would increase. B. The osmotic concentration of proteins in the dialysate fluid would increase. C. The osmotic concentration of proteins in the blood outflow would increase or remain unchanged. D. The filtration rate across the dialysis membrane would increase.

C. They contain mitochondria. One characteristic that distinguishes eukaryotic cells from prokaryotic cells is that eukaryotic cells contain membrane-bound organelles such as mitochondria.**

Which of the following characteristics clearly marks fungi as eukaryotes? A. They have cell walls. B. They contain ribosomes. C. They contain mitochondria. D. They exhibit sexual reproduction

B. High UV dewlap reflectance is most important in brightly lit habitats. Figure 1 shows that the three species of lizards that live in unshaded fields possess dewlaps that are significantly more capable of reflecting UV light than do the two species of lizards that live in the shaded understory. This supports the conclusion that high dewlap reflectance is most important in brightly lit habitats.** (passage says NOTHING about light)

Which of the following conclusions about dewlap reflectance is supported by information in the passage? (The five species are closely related and live in Puerto Rico. Three species (A, B, and C) live in open unshaded fields, and the other two species (D and E) live in the understory of a closed canopy forest.) A. Lizard habitat is determined by dewlap reflectance for each species. B. High UV dewlap reflectance is most important in brightly lit habitats. C. High dewlap reflectance is most important in dimly lit habitats. D. Dewlap reflectance is highest at the blue end of the visible spectrum.

D. 1. larger pores = faster filtration rate for equivalent weights (A or D) 2. higher molecular rate should be associated with low filtration (D correct)

Which of the following figures (A-D) shows expected solute filtration rates (mEq/mL-min) as a function of molecular weight for two dialysis membranes: Membrane 1 with large pores and Membrane 2 with small pores?

B.The digestive enzymes for the polysaccharide had to be transcribed and translated. the regulation of gene expression is one method by which bacteria respond to changes in their environment, in this case the addition of a new metabolite (polysaccharide). When polysaccharide is absent, the enzymes necessary to digest it are not expressed. However, upon addition of polysaccharide, in order for it to be used as an energy source, expression of the enzymes becomes necessary.***

Which of the following most likely explains why the bacterial colony did NOT grow immediately after the polysaccharide was added? A.The bacteria that were unable to digest the polysaccharide died. B.The digestive enzymes for the polysaccharide had to be transcribed and translated. C.The hydrolysis of fatty acids is a slow process. D.The polysaccharide directly inhibited bacterial fission

C. Enzymatic breakdown of food molecules The sympathetic nervous system directly inhibits peristalsis (A) and secretion of digestive enzymes (B). It also increases the blood glucose concentration and causes dilation of the blood vessels that supply the deep muscles and internal organs, which aids nutrient delivery (D) to these tissues. The sympathetic nervous system does not directly affect the activity of digestive enzymes (C) after they have been secreted**

Which of the following processes is LEAST directly influenced by adrenergic drugs? A. Peristalsis B. Secretion of digestive enzymes C. Enzymatic breakdown of food molecules D. Nutrient delivery to muscles and organs

D. The Regulative Hypothesis, because each embryonic cell receives a complete set of genes, and cell position helps to determine differentiation The Regulative Hypothesis fits our understanding of human development best and because cell position does have an effect on development in humans. ***

Which of the two hypotheses in the passage most closely fits the present-day understanding of human differentiation? A. The Mosaic Hypothesis, because each germ cell loses half of its genetic material during meiosis B. The Mosaic Hypothesis, because only specific genes are activated during the differentiation of each cell type C. The Regulative Hypothesis, because each embryonic cell receives a complete set of genes, and cell position is unrelated to differentiation D. The Regulative Hypothesis, because each embryonic cell receives a complete set of genes, and cell position helps to determine differentiation

D

Which primer is most suitable for PCR?

C. To maintain isotonicity of the dialysate solution with blood water is critical to survival so there should never be a correct answer where the goal is something to eliminate water from the body (therefore a and B wrong) also same values of Na indicate the preservation of isotonicity

Why are high concentrations of sodium included in the dialysate (Table 1)? A. To induce water movement from the blood into the dialysate fluid B. To maintain a high osmotic pressure in the dialysate solution C. To maintain isotonicity of the dialysate solution with blood D. To compensate for the urea nitrogen and creatinine in the blood

A. . No; aldosterone causes Na+ reabsorption by kidney tubules Because ingestion of excessive NaCl would trigger Na+ secretion into the urine, plasma-aldosterone levels would not increase. Rather, the body would rely on those homeostatic mechanisms that excreted the excess Na+. ** Thus an increase in plasma aldosterone would not be expected to follow ingestion of large quantities of NaCl.

Would an increase in the level of plasma aldosterone be expected to follow ingestion of excessive quantities of NaCl? A. No; aldosterone causes Na+ reabsorption by kidney tubules. B. No; aldosterone causes Na+ secretion by kidney tubules. C. Yes; aldosterone causes Na+ reabsorption by kidney tubules. D. Yes; aldosterone causes Na+ secretion by kidney tubules.

A.Veins, because the HCO3- concentration is higher in veins than in arteries.

Would the Cl- concentration of the RBCs be expected to be greater in the systemic veins or the systemic arteries? A.Veins, because the HCO3- concentration is higher in veins than in arteries. B.Veins, because there are fewer RBCs in veins than in arteries. C.Arteries, because the HCO3- concentration is higher in arteries than in veins. D.Arteries, because there are fewer RBCs in veins than in arteries