AAMC Biology Question Pack
What is the equation relating blood pressure, flow rate, and vascular resistance?
(VR): P = CO × VR.
Hemophilia, a disease in which the time required for blood to clot is greatly prolonged, is determined by a sex-linked gene. Suppose a man with normal blood clotting marries a woman with normal blood clotting whose father was a hemophiliac. If this couple has three sons, what is the probability that hemophilia will be transmitted to all three of them? A. 1/8 B. 1/4 C. 3/8 D. 1/2
A. 1/8 Sex-linked is X-linked. Father is normal so XAY and mother is a carrier (normal with one hemophilic allele) XAXa so they will produce sons that are: XAY and XaY meaning 50% will have hemophilia. For ALL three sons to have it, you would have to multiply the product o the three probabilities 1/2 x 3 = 1/8
Assuming that the vertebrates were all of comparable size, which of the following vertebrates would be expected to have the strongest and heaviest bones? A. A land-dwelling mammal B. A water-dwelling mammal C. A flying bird D. An amphibian
A. A land-dwelling mammal Lol, absolutely nothing to do with a recent common ancestor or anything. ALL THE SAME SIZE. A land-dwelling animal would be expected to have the heaviest bones (A), since enhanced bone density would be required to withstand the load bearing activity that results from the impact of gravity on land-dwelling animals. Less dense bones would be present in water-dwelling creatures (B) and amphibians (D), since the impact of gravity would be ameliorated by life in an aquatic environment. Likewise less dense bones would be a necessity for flight (C).
Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will both the + and - ends of the microfilament experience a net loss of subunits? A. At any concentration below 1 µM B. Exactly at 1 µM C. At any concentration above 1 µM D. Only between 1 µM and 4 µM
A. At any concentration below 1 µM Again, this is requiring close analysis of the figure! A "net loss" is experienced with there is more losing of actin than adding on both + and - ends. Because polymerization happens at a faster rate than depolymerization, anything above 1.5 µM (where tread-milling is occurring) is going to result in a net gain of actin. At below 1 µM, both the + and - end are experiencing depolymerization, which is a net loss.
Cancer cells most likely have an abnormality in their: A. DNA. B. rRNA. C. mitochondria. D. lysosomes.
A. DNA. Cancer cells are caused by mutations that go uncorrected in the DNA (are not corrected via DNA repair mechanisms and cell continues to replicate quickly).
Drugs affecting the ANS may have either an "active" mechanism (that mimics the activity of the PS or the S division) or a "passive" mechanism (that blocks the effect of the opposing division). The drug atropine prevents acetylcholine from attaching to its receptors on the circular muscle. Is this mechanism active or passive? A. Passive, because PS innervation is blocked B. Passive, because S innervation is blocked C. Active, because PS innervation is mimicked D. Active, because S innervation is mimicked
A. Passive, because PS innervation is blocked The question said that drugs affecting ANS are either "active" so agonists, or "passive" or antagonists. The passage mentions that acetylcholine is released at the motor synapse (at the organ) only in the PS division. Atropine prevents acetylcholine from attaching to receptor, so this is a PASSIVE mechanism that prevents PS innervation.
In the human "knee-jerk" reflex, the knee is struck and the lower leg jerks forward. Which of the following represents the complete pathway that the nerve impulse travels in effecting this response? A. Sensory neuron, motor neuron B. Sensory neuron, brain, motor neuron C. Sensory neuron, associative neuron, brain, associative neuron, motor neuron D. Sensory neuron, associative neuron, motor neuron, associative neuron, motor neuron
A. Sensory neuron, motor neuron
Why did investigators conclude that the new pathogen is a hantavirus? Experiment 1 Patients' sera were mixed with known pathogenic viruses and bacteria. A positive immunologic reaction was seen with a hantavirus that causes kidney disease. Experiment 2 Synthesized gene sequences from two known hantaviruses were mixed with nucleic acid from patients' lung tissues but not with lung tissue from control subjects. Hybridization between nucleic acid strands occurs when they base-pair with each other. Hybridization occurred between the synthesized genes and nucleic acid from the patients' lung tissues. Experiment 3 Genes from the new pathogen were inserted into bacteria to produce viral proteins, which were used to make pathogen-specific antibodies. These antibodies bound to capillary walls in patients' lung tissues. A. The experiments showed that it is related to known hantaviruses. B. The experiments showed that it infects lung endothelium. C. The patient data showed that the type of disease it caused resembles that caused by known hantaviruses. D. No pathogenic bacteria were found in the lung samples from patients.
A. The experiments showed that it is related to known hantaviruses. Experiment 1 is a clear example that what the patients had was similar to known hentaviruses, as shown the pathogenic antigens being similar to that of the hentavirus. Experiment 2 shows that the pathogen carries genes that are significantly similar in sequence to genes of two known hantaviruses. Taken together, these results support the conclusion that the pathogen is related to known hantaviruses (A). Because viruses other than hantaviruses can infect lung tissue and because known hantaviruses can infect tissue other than lung tissue, the observation that this pathogen infects lung tissue (B) does not demonstrate that the pathogen is a hantavirus. (C) is incorrect because the pathogen was infecting a different kind of tissue (lung) than the known hantavirus discussed in the passage was infecting (kidney). (D) is incorrect because the absence of pathogenic bacteria in infected tissue is not proof that the pathogen is necessarily a specific kind of virus.
Which result of Experiment 1 supports the hypothesis that cell-to-cell communication is involved in the determination of cell fate? A. The fate of an isolated AB cell differs from that of an AB cell in an intact embryo. B. The fate of an isolated P1 cell is indistinguishable from that of a P1 cell in an intact embryo. C. At the two-cell stage, isolated blastomeres can divide and differentiate. D. Several different blastomeres can produce both neurons and muscle tissue.
A. The fate of an isolated AB cell differs from that of an AB cell in an intact embryo. Carefully analyze the figure. Experiment 1 states: "To investigate the role of cell‑to-cell communication, researchers separated the cells of a two‑cell embryo and cultured them independently. The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo." The cells that are driving the differentiation are the ones that when cultured, produce everything, so P1! When P1 and AB are cultured separately, P1 cannot communicate to AB for it to differentiate into the appropriate tissues (missing muscle). If (B) was true, then this means that the P1 cell could develop tissues WITHOUT cell to cell communication (thus not proving the hypothesis and making it false! (C) and (D) are not telling us about direct cell-to-cell communication as it was being experimented on.
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.
A. fertilization and birth only. A female baby is born with all the eggs, primary oocytes, she will need in her lifetime. These primary oocytes are arrested in prophase I at the time of birth. So they have already gone though mitosis and have started meiosis I. At puberty, each moth, some of these oocytes complete meiosis I and start meiosis II. When they are ovulated, they are arrested in metaphase II. Meiosis II completes only when fertilization takes place but the fusion of nuclei of the egg and sperm takes place once meiosis II completes. So basically, Embryo forms -> Gonads differentiate to ovaries (female) -> germ cells differentiate to oogonia --> MITOSIS I and MEIOSIS II of primary oocytes that are arrested at Prophase 1 (millions) --> ovulation leads to secondary oocytes arrested at METAPHASE II, which is finally completed after fertilization before fusion of sperm and egg nuclei.
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.
A. glomerular filtration rate. The best answer is that increased blood pressure will affect the glomerular filtration rate, answer choice A. Tm is a characteristic that depends on the characteristics of the cells lining the renal tubules and independent of blood pressure, so answer choice B is not correct. Water resorption and concentrating ability are the same, so answer choices C and D are essentially the same. Increasing blood pressure should increase flow of fluid through the kidney system and decrease, rather than increase, water reabsorption, so these answer choices are incorrect.
During the initial skin diving session, when her heart and breathing rates were increased, Sarah noticed that she produced more urine than usual. This was most probably a result of: A. increased blood pressure caused by her excitement or anxiety. B. reduced blood pressure caused by her excitement or anxiety. C. absorption of water from the ocean. D. inability to cool the skin through evaporative water loss.
A. increased blood pressure caused by her excitement or anxiety. Higher blood pressure means more kidney is producing more filtrate and excreting more water!
Applying a drug which blocks the absorption of NE into the adrenergic nerve terminal will result in: A. increased sympathetic responses. B. decreased sympathetic responses. C. increased destruction of NE by MAO. D. decreased destruction of NE by COMT.
A. increased sympathetic responses. The re-absorption of NE into the synaptic terminal from the synaptic cleft is what decreases whatever effect it is causing. In this cause, NE in the synaptic cleft is activating the sympathetic nervous system. A drug that blocks re-absorption into adrenergic nerves will increase the sympathetic response! For example, a SSRI is something that inhibits the re-uptake of serotonin to INCREASE serotonin in the synaptic cleft for a heightened effect.
A wound that penetrates the rib cage and lets air into the right pleural cavity stops airflow into the right lung because the: A. lung cannot be expanded. B. rib cage cannot be expanded. C. diaphragm cannot be lowered. D. air dries and stiffens the lung.
A. lung cannot be expanded. Non-forced inspiration occurs when the diaphragm and muscles between the ribs (intercostal muscles) contract. The ribs attached at the sternum and spine lift in what is called a "bucket-handle" motion and the diaphragm flattens. These motions increase the volume of the thoracic cavity. Between the lung and the thoracic wall is a space, the pleural cavity. As the volume of the thoracic cavity increases, atmospheric pressure presses against the inside of the flexible, elastic lung tissue forcing it to conform to the thoracic wall. Air thus enters the lung as it expands with the thorax. If the thoracic wall is pierced (causing what is known as a sucking chest wound), air enters the pleural cavity. The air pressure inside and outside the lung balance causing the lung to elastically collapse and preventing air intake. The rib cage can still expand (so choice B is incorrect), the diaphragm can still contract (so choice C is incorrect), but the lung cannot be expanded in response, so the correct response is A.
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. more negative IPP and inspiration. Remember: Inhaling is due to a negative pressure system. The diaphragm is a muscular partition between the abdominal and thoracic cavities. It is dome-shaped at rest curving up toward the lungs and heart. It flattens when it contracts during inspiration. Because it is anchored around its edges to the ribs and spine, when the diaphragm contracts the volume of the thoracic cavity increases. This decreases the pressure (IPP) in the pleural cavity between the thoracic wall and the lung. Fresh air flows in to equalize the pressure inflating the lung.
The osmotic concentration of plasma proteins in the venous side of capillaries helps reduce the amount of interstitial fluid in tissues by inducing: A. passive H2O diffusion along a concentration gradient. B. passive ion diffusion along an electrochemical gradient. C. facilitated ion transport along an electrochemical gradient. D. active H2O transport mediated by an ATP-dependent pump.
A. passive H2O diffusion along a concentration gradient. The greater osmotic concentration of plasma proteins on the venous side will cause osmotic pressure to "suck" water back in from the interstitial tissues, thus decreasing fluid content in the interstitial tissues. Water is higher in the tissues than the venous side of the capillaries, so it will go down its concentration gradient. Water is lower in the tissues on the arteriole side of the capillaries, so hydrostatic pressure will cause fluid to go INTO the interstitial tissues (starling forces). (B) and (C) don't describe fluid movement, they are related to solute concentration and osmotic concentration is governed by fluid flow! Water is never transported with an ATP-pump. There are only two main ways that water is transported: -Water crosses cell membranes by passive transport down its concentration gradient or by secondary active co-transport along with ions.
One characteristic common to arteries, veins, and capillaries is the: A. presence of a layer of endothelial cells. B. presence of numerous valves that prevent the backflow of blood. C. ability to actively dilate or constrict in regulating blood flow. D. ability to supply surrounding tissues with nutrients by filtration and diffusion.
A. presence of a layer of endothelial cells. The question asks the examinee to identify one characteristic common to arteries, capillaries, and veins. A is correct because all three types of vessels possess an inner layer of endothelial cells. B is incorrect because only veins have valves. C is incorrect because only certain types of arteries dilate or constrict to regulate blood flow. D is incorrect because the exchange of nutrients with the surrounding tissues occurs only in capillaries.
Tests performed on the M. tuberculosis strain infecting the patient's coworker indicated that the strain was susceptible to both ampicillin and kanamycin, and the coworker was successfully treated. The M. tuberculosis most likely survived in the patient because it had: A. undergone conjugation with cells of resistant E. coli. B. undergone an antibiotic-induced mutation that conferred antibiotic resistance. C. reproduced more rapidly than the strain in the coworker. D. adapted to its new environment by modifying its metabolism.
A. undergone conjugation with cells of resistant E. coli. This really should have been a no-brainer. ANTIBIOTICS DO NOT INDUCE MUTATIONS. Antibiotic-resistant bacteria develop when mutations occur in their genome that makes them resistant (while non-resistant ones are killed off) and they pass this on during cell division OR through HGT (conjugation, transposition, or transformation). (D) and (C) do not explain why the antibiotic was ineffective- why would it reproduce more rapidly in one environment over the other? The M. tuberculosis did not adapt to its new environment by modifying its metabolism (D); rather, there was a strain with a metabolic capability that was not compromised by the antibiotic
What is the difference between ADH and Aldosterone in terms of regulating BP?
ADH works on the collecting duct (by increasing permeability of collecting duct to water through bringing more aquaporins to surface of cells). It DOES affect blood osmolarity because all it does is increase water (and thus, decrease solute concentration). This decreases blood osmolarity. Aldosterone works on both the collecting duct and the DCT. It does not affect blood osmolarity since it increases solute AND water levels (pulling in both Na+ and H2O, as you mentioned).
Describe sequence of channel openings/closings in action potential.
Action potential generally follows a pattern. 1. A cell's resting potential is at -70mV -Negative due to leak of potassium ions out of the cell 2. When a stimulus is applied, the membrane potential is increased. If a threshold membrane potential is reached, the sodium channels will open 3. Once the sodium voltage gated channels have opened, sodium will diffuse into the cell, raising the membrane potential to ~40mV. -When measuring action potential, this creates a negative current 4. After a period of time, the sodium channels will inactivate, causing the flow of sodium to stop 5. Potassium voltage gated channels will then open, causing potassium to diffuse out of the cell, causing hyperpolarization, after which K+ channels close and the and leak channels works to bring axon back to resting membrane potential -back to -70mV. This is assisted by chlorine channels opening, resulting in chlorine diffusing into the cell. -When measuring action potential, this creates a positive current -The membrane potential may drop slightly below resting potential (hyperpolarization). At this point all potassium channels close.
What is the role of ADH and Aldosterone in water re-absorption?
Aldosterone (INCREASE BP) -The need to conserve water in the body would lead to an increase in aldosterone secretion that would promote the retention of sodium ions. The retention of sodium ions would in turn promote the osmotic reabsorption of water by the kidneys. -no change in blood osmolarity ADH (INCREASE BP) -Antidiuretic hormone (ADH or vasopressin) secretion would increase to enhance the permeability of the distal tubules and collecting ducts of the kidney to promote water reabsorption and excretion of more concentrated urine. -Make leaky cell junctions -Decrease blood osmolarity
Where are the anterior, posterior, and lateral canals located?
At the beginning of the semicircular canals after the bony labyrinth. Give information about head rotation. Endolymph lags in the ampulla when the canal wall and cupula rotate, causing endolymph to exert force on cupula causing it and hair cells to bend (depolarize/hyperpolarize).
Consider an organism that has three pairs of chromosomes, AaBbCc, in its diploid cells. How many genotypically different kinds of haploid cells can it produce? A. 4 B. 8 C. 16 D. 32
B. 8 The number of different possible gametes that can be formed by diploid organisms as a result of independent assortment of chromosomes during meiosis can be calculated using the formula 2^n where n is the haploid number of chromosomes. In this case, the haploid number is 3, making the number of different haploid cells 23, or 8. Thus, B is the best answer.
What was contained in the sera from the respiratory patients of Experiment 1 that caused the sera to react with a hantavirus that causes kidney disease? Experiment 1 Patients' sera were mixed with known pathogenic viruses and bacteria. A positive immunologic reaction was seen with a hantavirus that causes kidney disease. A. Antibodies specific for the kidney disease hantavirus only B. Antibodies to the unknown pathogen, which is antigenically related to the known hantavirus C. Antibodies to kidney proteins D. The unknown pathogen
B. Antibodies to the unknown pathogen, which is antigenically related to the known hantavirus. Firstly, it is important to understand what a sera is. The sera is the blood serum of an animal, used especially to provide immunity to a pathogen or toxin by inoculation or as a diagnostic agent. The researchers took the patient's sera (the ones who had the pathogen affecting their lungs), and mixed with other pathogenic viruses (known) and saw that there was a positive response for kidney disease hentavirus--this means that the patients must have had developed antibodies to the the unknown pathogen which is similar to that of hentavirus causing kidney disease.
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.
B. EMS to E. READ THE QUESTION CAREFULLY (AND LOOK AT APPROPRIATE FIGURES). **not asking about cell-to-cell communication, but cell content segregation! This flow chart gives you a really straightforward answer as to how the differentiation/segregation occurs. I was looking at the second figure, but it is important to see that P1 directly gives rise to EMS and P2 at the 4-cell stage and then the appropriate tissues. If the zygote has all of the necessary contents for differentiation, segregation of the substances required for gut differentiation would happen from zygote -> P1 -> EMS -> E. I was getting this confused with cell-to-cell communication in the 4-cell stage from the previous question. EMS still needs the cell contents from P1 and then cell-to-cell communication from P2.
Which of the following hormones is LEAST directly regulated by the anterior pituitary? A. Cortisone B. Epinephrine C. Progesterone D. Thyroxin
B. Epinephrine Epinephrine is secreted by the adrenal medulla and its secretion is controlled by the autonomic nervous system. Secretion of the other three hormones is regulated by secretion of anterior pituitary hormones. Cortisone secretion is regulated by adrenocorticotropic hormone (ACTH), progesterone by luteinizing hormone (LH), and thyroxin by thyroid stimulating hormone (TSH)
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
B. Maintain sufficient oxygenation of cells She is in a cold environment. Her vessels are probably constricted to prevent heat loss (thermoregulation). However, if they remain constricted for too long, her cells will not get the oxygenation they need, to it will occasionally dilate to allow the cells to receive more oxygen.
The Ames test is often used in the initial screening of suspected carcinogenic compounds because it provides a good indication of the mutagenic characteristics of many chemicals. The test uses special strains of the bacterium Salmonella typhimurium that are nutritional mutants; they also lack a mechanism for DNA repair. When grown on a medium that lacks the amino acid histidine, the Salmonella test strains do not survive even though wild-type Salmonella grow well on this medium. During the Ames test, the suspected carcinogen is added to a histidine-deficient growth medium. If the chemical is a mutagen, some of the bacteria will back-mutate, and a visible colony will form. The usefulness of the Ames test can be improved when the growth medium contains rat-liver enzymes. Why is the Ames test for mutagens used to test for carcinogens? A. Salmonella transform mutagens into carcinogens. B. Most mutagens are also carcinogens. C. Salmonella contain oncogenes. D. Salmonella's RNA distinguishes between carcinogens and mutagens.
B. Most mutagens are also carcinogens. The Salmonella Typhimurium is a nutritional mutant that lacks a mechanism for DNA repair. This means that adding a carcinogen to the bacteria will mutate the Salmonella (not vice versa!). B. Most mutagens are also carcinogens.
The results of Experiment 1 indicate that the direction of signaling between the blastomeres of a two-cell embryo is: A. AB → P1. B. P1 → AB. C. P1 → P2. D zygote → AB.
B. P1 → AB. In Experiment 1, researchers investigated the role of cell-to-cell communication by separating the cells of a two-cell embryo (AB and P1), and culturing them independently. The question stem is asking about the BLASTOMERES of the TWO-CELL embryo! So we can eliminate (D) because they zygote is not the blastomere as well as (C) because P1 is part of the two-cell, while P2 is part of the 4-cell blastomere. To decide between (A) and (B), we look to the rest of the experiment: The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo. The results of Experiment 1 indicate that the direction of signaling between the blastomeres of a two-cell embryo is P1 → AB (since AB's fate differed when isolated from P1, while P1's fate was the same regardless of whether AB was present or not).
Radioactively labeled uracil is added to a culture of actively dividing mammalian cells. In which of the following cell structures will the uracil be incorporated? A. Chromosomes B. Ribosomes C. Lysosomes D. Nuclear membrane
B. Ribosomes Don't rush! Think about what this question is saying. Uracil is specific to RNA! So it wouldn't be found in the chromosomes which is DNA. Ribosomes is where mRNA is translated, so that would be a place with radioactively labeled Uracil would be incorporated. If it said nucleus, that would also be a possibility since transcription occurs in the nucleus.
Equal concentrations of 8 mg/mL of Substance A and glucose are found in a volunteer's plasma. Based on Figure 1, which substance will the kidney clear from the plasma more rapidly? A. Substance A, because the slope of the clearance line for Substance A is higher than that for glucose B. Substance A, because Substance A reaches its Tm at a lower plasma concentration than does glucose C. Glucose, because glucose reaches its Tm at a higher plasma concentration than does Substance A D. Glucose, because the slope of the clearance line for glucose is lower than that for Substance A
B. Substance A, because Substance A reaches its Tm at a lower plasma concentration than does glucose The Tm is a characteristic of the individual substances in the tubule system and a measure of how efficiently each substance can be reabsorbed. A high Tm indicates a high capacity for reabsorption of substances in the kidney tubules. In figure 1, the Tm for each substance can be read as the concentration in plasma when the concentration in the urine is zero. In this case it looks like the Tm for Substance A is a little over 6 mg/mL and that of glucose is 10 mg/mL. So substance A has a lower Tm. This means the tubules will not reabsorb it very efficiently. Much of it will be spilling into the urine, thus being eliminated from the body. The question asks which will clear from the blood more rapidly at a concentration of 8 mg/mL and why. The answer is that all glucose will be reabsorbed at that concentration, none will appear in the urine and none will be cleared from the plasma. Glucose has such a high Tm, that all of the glucose will be reabsorbed into the bloodstream, perhaps to reenter the kidney tubule again. A rate of 8mg/mL is above the Tm of substance A, so there will be some substance A in the urine at this plasma concentration. The answer to the question depends on the value of Tm, not the slope of the clearance line, so answer choices A and D can be eliminated. Substance A will clear more rapidly than glucose, therefore, answer choice B is correct
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.
B. The - ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. The question stem is saying that the microfilament length DOES Not change when the sarcomere shortens in muscle contraction. This means that if one end (the - end) of the microfilament is capped by Z-lines (to prevent depolymerization--losing actin subunits), then the positive end also has to be "prevented" from polymerizing. The passage mentions "capping" proteins that regulate microfilament growth. From the graph, we can see that above 1 µM, the positive end of the actin subunits is experiencing addition, so (A) and (C) can't be true.' (D) is not plausible because the figure is showing that the microfilament is being stopped from addition or subtraction by the Z-line.
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
B. The coordination of cell differentiation during development. Differentiation depends on the suppression and activation of different genes, including genes that have already been expressed. If a gene has expressed RNA before being silenced, that mRNA is still in the cytoplasm, creating proteins. If it is not degraded, it will continue to produce those proteins and prevent differentiation. For example, in the differentiation of fingers, a gradient is formed by cells using Shh proteins. Cells that will differentiate into the thumb will have a high Shh concentration, cells that will differentiate into the index finger will have less, etc. etc., and the cells determined for pinky will have none. A big part to this relies on suppressing mRNA so that Shh protein is not expressed in the pinky. If mRNA is not suppressed, the cells cannot be determined and differentiation will be consequently adversely affected.
Which of the following statements correctly describes the distinction between the exocrine and endocrine portions of the testis? A. The exocrine portion secretes only peptides; the endocrine portion secretes only steroids. B. The exocrine portion releases its products into ducts; the endocrine portion releases its products into the blood. C. The exocrine portion secretes only cellular elements; the endocrine portion secretes only chemical substances. D. The exocrine portion is the target tissue for the products of the endocrine portion.
B. The exocrine portion releases its products into ducts; the endocrine portion releases its products into the blood. Endocrine is a form of cell-cell communication that involves the secretion of hormones into the bloodstream by ductless glands, and these hormones go on to travel to distant locations within the organism to cause a change in cellular activity. Exocrine are glands that release their secretions into ducts (such as parts of the liver and sweat glands).
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
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. Thus, B is the best answer.
All of the following occur during normal inspiration of air in mammals EXCEPT: A. elevation of the rib cage. B. relaxation of the diaphragm. C. reduction of pressure in the pleural cavity. D. contraction of the external intercostal rib muscles.
B. relaxation of the diaphragm. Remember. INHALING -Increase volume -Intercostal and diaphragm muscles contract -Alveoli are stretched (elastin), increase volume EXHALING -Decrease volume -Intercostal and diaphragm muscles relax -Alveoli "recoil", decrease volume
All of the following are functions of mammalian skin EXCEPT: A. sensation. B. respiration. C. protection from disease. D. protection against internal injury.
B. respiration. LMFAOOOO respiration is the LUNGS sis.
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. the peritoneal cavity. (A) the perineum is the the area between the anus and the scrotum or vulva. (B) The peritoneal cavity is a potential space between the parietal peritoneum (the peritoneum that surrounds the abdominal wall) and visceral peritoneum (the peritoneum that surrounds the internal organs). (C) the pleural cavity is the potential space between the two pleurae (visceral and parietal) of the lungs (D) the lumen is where the contents of the GI are original before the ulcer penetrated.
What is the process of spermatogenesis from a spermatogonium to a mature spermatozoa?
Before puberty, the seminiferous tubules contain only spermatogonia and Sertoli cells. Beginning at puberty, each spermatogonium will undergo a series of mitotic and meiotic divisions, called spermatogenesis, that result in the production of mature spermatozoa (analogous to female ovum). 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 Sertoli, or "nurse" cells, provide nutrients for the developing sperm. In addition, the Sertoli cell membranes form tight junctions, establishing a blood-testis barrier that protects developing sperm from potentially toxic bloodborne substances, such as proteins and polar compounds.
Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will the microfilament treadmill? A. 0.25 µM B. 1.0 µM C. 1.5 µM D. Any concentration between 1.0 µM and 4.0 µM
C. 1.5 µM Analyze the figure carefully!!! Tread-milling is defined by equal addition and subtraction of actin. Look on the graph at the point when + end of actin is adding and - end of actin is subtracting. Until 1 µM, it is only subtracting, so (A) and (B) can be eliminated. At 1.5 µM, the + end is two away from equilibrium for addition, and the - end is two away from equilibrium for subtraction. Because the rate of addition is greater than the rate of subtraction, it cannot be treadmilling at any concentration between 1-4, eliminating (D) as well.
Suppose that in a randomly mating population of mammals, 160 of its 1,000 members exhibit a specific recessive trait that does not affect viability of the individual. How many individuals in this population are heterozygous carriers of the gene that causes this trait? A. 160 B. 400 C. 480 D. 600
C. 480 The gene frequency of the recessive trait, q, can be ascertained by taking the square root of 16/100 or 0.4. p + q = 1 p^2 + 2pq + q^2 = 1 Given this frequency, the frequency of p must be 1.0 - 0.4 = 0.6. According to the Hardy-Weinberg law, the frequency of heterozygous carriers is thus given by 2 pq or 2 X .4 X .6 = .48 or 48%. 48% of 1000 = 480. Thus, answer choice C is the correct answer.
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, what is the net volume of fresh air that enters the alveoli each minute? A. 650 mL B. 785 mL C. 6,500 mL D. 7,850 mL
C. 6,500 mL The tidal volume is the amount of air that moves into the lungs during each inspiration (800 mL) Some of this air does not reach the alveoli so it is not available for gas exchange with the circulatory system. The volume of such gas in the air passageways, (trachea, bronchi and bronchioles) is called the respiratory dead space (150 mL). Subtracting the dead space of 150 mL from the tidal volume of 800 mL gives 650 mL, the quantity of air that enters the alveoli on each inspiration. Over ten breaths then, 6,500 mL will flow into the alveoli for gas exchange (650 mL x 10 breaths).
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.
C. E. coli entering the abdominal cavity from the appendix. APPENDIX IS CONNECTED TO COLON. If it belongs in one, it belongs in the other. Bacterial infections happen when something goes where it ain't supposed to be at :) 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.
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
C. Enzymatic breakdown of food molecules Adrenergic drugs are responsible for activating the sympathetic nervous system. The sympathetic nervous system directly inhibits peristalsis and secretion of digestive enzymes. 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.
The amount of NE released by sympathetic nerve terminals will be most strongly influenced by a change in which of the following? A. Alpha receptor sensitivity B. Alpha receptor density C. Extracellular [Ca2+] D. COMT activity
C. Extracellular [Ca2+] V-gated Ca2+ channels respond to depolarization once it reaches synaptic terminal and allow Ca2+ to come in which causes vesicles to fuse with membrane and release neurotransmitter. Receptors on the post-synaptic terminal are RECEIVING NE, NOT RELEASING NE. So (A) and (B) wouldn't impact the release itself, only the response to NE release. COMT activity happens in the synaptic cleft, so post-release of NE from pre-synaptic terminal into the cleft.
Hypothesis A An analog of Ohm's Law applied to the cardiovascular system illustrates the basic relationship between blood pressure (P), flow rate of blood from the heart (cardiac output or CO), and vascular resistance to the flow of blood (VR): P = CO × VR. Vascular resistance is predominantly a function of blood vessel radius. Therefore, an increase in systemic vascular resistance caused by factors such as vessel disease or enhanced muscle tone in the vessel walls (vasoconstriction) may be the major cause of systemic hypertension. Hypothesis B Although increases in resistance to blood flow can quickly increase blood pressure, increased pressure should presently act to initiate an effective corrective reflex involving the kidneys. The increased pressure should cause the kidneys to increase their output of fluid, and this should bring the pressure back to normal despite the persistent elevation in vascular resistance. The nervous system is probably not involved in this reflex. Failure of this reflex function may cause systemic hypertension. 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
C. Hypothesis B, because the kidneys were responding to decreased glomerular blood pressure If a renal artery is clamped, this means that less blood is flowing into the glomerulus, so the glomerulus is filtering less blood, resulting in lower excretion of fluid. This will cause kidneys to resaborp more fluid to increase BP. Hypothesis A is mainly dealing with systemic hypertension that comes from reduced vessel radius (vasoconstriction) or decreased muscle tone. These all result in increased vascular resistance and increased blood pressure. However, it is not addressing specifically the role of the kidney, which hypothesis B is doing by talking specifically about the homeostatic function of the kidneys in regulating BP (we are putting a RENAL arterial clamp after all.
Which participant in the electron transport chain has the greatest attraction for electrons? A. FAD B. NAD+ C. Oxygen D. Cytochrome c
C. Oxygen Electrons move to a slightly more electronegative carrier as the electrons pass through each step in the electron transport chain. Oxygen is the final electron acceptor in the ETC, therefore it has to have the highest e- affinity (greatest E/N) because it is not "giving up" its electrons to anything. (A) FAD and (B) NAD+ do not participate in the ETC, they are the products of NADH and FAHD2 reduction at Complex I and II. (D) is what I chose but wrong because Cyt C is also an e- donor to Complex IV.
If an artery that supplies blood to a lung lobe was blocked but ventilation to the lobe was unaffected, how would alveolar gas partial pressures change? A. Both PO2 and PCO2 would increase. B. Both PO2 and PCO2 would decrease. C. PO2 would increase and PCO2 would decrease. D. PO2 would decrease and PCO2 would increase.
C. PO2 would increase and PCO2 would decrease If the blood flow to an alveolus were blocked there would be no flow of hemoglobin-rich red blood cells to take away O2 and no influx of CO2 from the blood. As a result the air in the alveolus would become more like that of the atmosphere. It would acquire a higher PO2 and a lower PCO2.
Which of the following best explains why bacterial colonies formed on Plate IV in Figure 1? A. The air contained mutagens. B. The agar contained mutagens. C. Spontaneous mutations occurred. D. The DNA repair system became activated.
C. Spontaneous mutations occurred. The question asks the examinee to explain why Plate IV - which was exposed to purified air and no mutagens - had some colonies growing on it. T The most likely explanation for the presence of these colonies is that they are due to spontaneous mutations (C) in the DNA of the Salmonella. One can assume, based on information provided in the passage, that the air (A) and media (B) to which the bacteria are exposed are pure and free of mutagens, and thus not causing any mutations in the test strains. If DNA repair systems were activated any back-mutations that occurred would be corrected and no colonies would form on the plates (D).
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
C. To maintain isotonicity of the dialysate solution with blood The table shows ranges of solute concentrations in normal plasma, renal failure plasma, and the dialysate fluid. The question is asking about sodium concentrations int he dialysate as compared to the normal/renal---because there is no change in the sodium concentrations between dialysate and normal/renal failure, (A) and (B) cannot be true. (D) is just false because sodium concentration has nothing to do with urea or creatinine in the blood :)
Plasma clearance is affected by the tubular transport maximum (Tm) of a substance. The Tm is the maximum rate of transport (mg/min) at which a substance can be reabsorbed by the kidney. That is, if the filtration rate of a substance exceeds its Tm, the substance will begin to appear in the urine. The Tm for glucose averages 320 mg/min in an adult human. According to the passage, the Tm represents the rate of plasma filtration that just exceeds the: A. rate of concentration of the substance in the glomerular filtrate. B. rate of concentration of the substance in the urine. C. capacity of the kidney tubules to reabsorb the substance. D. capacity of the bladder to store and excrete the substance.
C. capacity of the kidney tubules to reabsorb the substance. The glomerulus is a tuft of capillaries that bulges into the capsular space (also referred to in some texts by its eponym, Bowman's space), which is a potential space lined with simple squamous epithelium. Fluid is expressed from the blood across the capillary endothelium. It enters first the capsular space, then the kidney tubule. The part of the tubule closest to the glomerulus is the proximal tubule, the next part is the U-shaped loop of Henle and the last part of the tubule is the distal tubule. Suppose one gradually increased the rate at which fluid was expressed from the bloodstream (the glomerular filtrate rate) and measured the concentration of some substance reabsorbed by the kidney as it left the distal tubule as urine. The concentration might not rise at first, because the cells lining the tubule might be completely reabsorbing the substance and putting it back into the blood stream. Eventually, however, the rate of flow would reach a point at which it exceeds the rate at which the tubule cells could reabsorb the substance. The fluid would be flowing too rapidly through the tubule for the cells to reabsorb all the substance. The rate of flow through the tubule at which the substance begins to be observed in the urine is Tm. At that point the rate of flow of fluid through the tubule begins to exceed the capacity of the kidney tubule cells to reabsorb the substance
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. meiosis. The macronucleus does not go through the process of meiosis. Genes involved in meiosis are therefore superfluous in this genome. The correct answer is choice C, that DNA sequences involved in meiosis may be eliminated. On the other hand the primary use of the macronucleus is to provide the proteins for the cell's day-to-day functioning. 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.
The mineral component of human bone is a salt that consists primarily of all of the following EXCEPT: A. calcium. B. phosphate. C. potassium. D. hydroxyl groups.
C. potassium. The inorganic component of bone consists of submicroscopic deposits of calcium phosphate similar to hydroxyapatite (Ca10[PO4]6[OH]2), so it would be expected to contain calcium, phosphate and hydroxyl groups.
Fats are known to affect blood flow...Vitamin E, an effective anti-oxidant, was given to 2 groups to reduce in vivo oxidation of the ingested fatty acids. On the back! 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. the products of fatty acid oxidation reduce skin blood flow. We can eliminate options (A) and (B) because vitamin E was never tested alone, but in the cases that it is added to fatty acid and placebo, the blood flow rate increases. Since Vitamin E (which prevents the in vivo oxidation of FA) is shown to increase blood flow, and FA alone are shown to reduce blood flow, this must mean the products of FA that are oxidized are reducing blood flow.
Where is blood pressure the lowest: Heart Arteries Arterioles Capillaries
Capillaries
Where does transcription occur in prokaryotes?
Cytoplasm (translation immediately follows)- these processes are occurring simultaneously
After Sarah's accident, her attending physician detected the protein myoglobin in her urine. What type of injury is consistent with this observation? Broken bone Damaged muscle Damaged kidney A. I only B. III only C. I and III only D. II and III only
D. II and III only Myoglobin is only found in the muscle. It is not present in the bloodstream, or really anywhere else in the body. Increased myoglobin presence in the blood indicates a muscle injury. The fact that it is found in the urine means that the nephrons are not filtering out proteins properly (proteins are normally filtered in the membranes of the glumerulus, where the basolateral membrane prevents the passage of proteins into the urine). The injury is likely both in the kidney and muscle.
What mechanism probably would be responsible for the increased urine output induced by hypertension according to Hypothesis B? A. Increased blood flow to the bladder B. Increased renal tubular reabsorption of solutes and water C. Increased collecting duct permeability to water D. Increased glomerular filtration rate
D. Increased glomerular filtration rate The question is asking about INCREASED urine output. (B) and (C) would both result in DECREASED urine output because more water would be resorbed from the DCT and CD into the interstitial space. (D) is the answer because the increased pressure should cause the kidneys to increase their rate of filtration to increase the output of fluid. Increased blood flow to the bladder would increase bladder pressure but not necessarily effect the urine output.
Increased vasoconstriction has an important role in which of the following situations? A. Causing the decrease in blood pressure associated with fainting B. Increasing blood flow to muscle during exercise C. Increasing blood flow to skin during blushing D. Maintaining blood pressure during a hemorrhage
D. Maintaining blood pressure during a hemorrhage (A) caused by DILATION-Though out of scope, Vasovagal Syncope (common fainting) is due to the blood vessels in your legs dilating. This allows blood to pool in your legs, which lowers your blood pressure. Combined, the drop in blood pressure and slowed heart rate quickly reduce blood flow to your brain, and you faint. (B) and (C) are also caused by vasodilation.
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
D. No, because blockage of phenylalanine hydroxylase gene expression will not remedy the original disorder. This is a mutation of a gene that causes inhibited expression. Antisense genes inhibit expression by preventing translation, but are not useful for increasing expression.
Production of which of the following hormones will be inhibited by the administration of dietary calcium to prevent osteoporosis? A. Growth hormone B. Calcitonin C. Thyroid hormone D. Parathyroid hormone
D. Parathyroid hormone When blood calcium levels are HIGH, calcitonin is secreted by the thyroid gland for Ca2+ deposits onto bones and reduction of Ca2+ uptake into tissues. Parathyroid hormone activity is INHIBITED. When blood calcium levels are LOW, parathyroid gland senses this and releases parathyroid hormone (PTH) to activate Vitamin D and increase Ca2+ uptake into kidney and intestines, and increase Ca2+ release from bones. INHIBIT CALCITONIN.
The results of Experiment 2 indicate that the signaling interaction at the two-cell stage probably most involves which class of macromolecules? A. DNA B. Messenger RNA C. Ribosomal RNA D. Protein
D. Protein The AB cells had their translation inhibited and were not able to produce all of the tissues of the original embryo, but were still able to produce tissue when transcription was inhibited. Even though you've inhibited transcription and translation at this point that doesn't mean the two cells are bereft of mRNA/proteins. So what happens when you halt translation is that all currently made mRNA can't become protein. If you halt transcription you just can't make more mRNA but you can translate all the current mRNA into protein. In Experiment 2, two-cell embryos were incubated in either cycloheximide (an inhibitor of translation) or actinomycin D (an inhibitor of transcription). The AB cells were then isolated and washed to remove inhibitors, and grown in culture. AB cells of embryos treated with cycloheximide (the translation inhibitor, which would have prevented the production of proteins at the ribosomes of both AB and P1 cells) produced only neurons and skin. AB cells of embryos treated with actinomycin D (the transcription inhibitor, which would have prevented production of mRNA) produced neurons, skin and muscle—their normal fate. These results indicate that the signaling interaction (between P1 and AB cells) at the two-cell stage probably involves protein, since proteins of the P1 cells could not have been produced to carry the necessary message(s) to the AB cells prior to isolation.
If acetylcholine is removed from the circulation faster than norepinephrine is, which of the following autonomic processes would be most rapidly inactivated? A. Dilation of the pupils B. Dilation of blood vessels in the skeletal muscles C. Rise in blood pressure D. Stimulation of digestive secretions
D. Stimulation of digestive secretions All of the other options are caused by Sympathetic (norepinephrine) activation: -Dilated pupils -Dilation of blood vessels in skeletal muscles, constriction of blood vessels everywhere else -Rise in BP -Restriction of peristalsis and digestive secretions If acetylcholine is REMOVED, and there is still NOREPINEPHRINE, the PARASYMPATHETIC NERVOUS SYSTEM is going to be inactivated, so no stimulation of digestive secretions.
If a cell's membrane potential changes from -60mV to -70mV after treatment with an adrenergic drug, the NE receptor is most likely linked to: A. a G-protein. B. adenylate cyclase. C. a sodium channel. D. a potassium channel.
D. a potassium channel.
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.
D. a variable allele distribution in the macronucleus. The heterozygous macronucleus is a diploid, germ-line micronucleus and 45-ploid. This means that it will have a variable allele distribution (D). It's copies of 45 chromosomes (so uneven!). When you have an odd number in the copies stage, you end up with anomalies. The passage states that the old macronucleus becomes the new micronucleus, so (C) is false and can be ruled out. Crossing over only happens in meiosis and the macronucleus undergoes amitotic division based on passage information, so crossing over cannot occur! The passage also states that each daughter cell receives equal amount of DNA from the amitotic nucleus so binary fission wouldn't contribute to a (A) a loss of macronuclear chromosomes.
Inbreeding can reduce the fitness of a population in the short term because it causes an increase in the: A. genetic diversity of the population. B. levels of aggression in the population. C. rate of spontaneous mutations. D. incidence of expression of deleterious recessive traits.
D. incidence of expression of deleterious recessive traits Deleterious genes are generally rare because they tend to be eliminated through natural selection. Only when an organism is homozygous (has two copies of a gene, one from each parent) does a recessive gene reveal its presence. Because recessive genes can be masked by dominant genes, they are less exposed to natural selection. So most organisms carry many deleterious recessive genes. The chances of having offspring that are homozygous for a given recessive gene are rare when a mate is chosen randomly from the population. The chances of getting a pair of deleterious recessive genes increase enormously when the mate is a relative, because relatives are likely to have a similar genotype. The answer is therefore choice D. Inbreeding decreases rather than increases genetic diversity (choice A), its effect on aggression is hard to determine (Choice B) and it has no effect on spontaneous mutations (choice C), so these are incorrect choices.
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.
D. it suppresses cells specific to the body's own antigens. Self-antigens are proteins and carbs present o the surface of every cell of the body. Under normal circumstances, these self-antigens signal to immune cells that the cell is not foreign and should not be attacked. However, when the immune system fails to make the distinction between self and foreign, it may attack cells expressing particular self-antigens a condition known as AUTO-IMMUNITY. The human body strives to prevent autoimmune diseases very early in T-cell and B-cell maturation. Negative selection: B-cells maturate in the bone marrow and ones that respond to self-antigens are eliminated before they can leave the bone marrow. T-cells maturate in the thymus and ones that respond to self-antigens are eliminated by Suppressor T cells---promoting SELF-TOLERANCE.
The fact that there appears to be a genetic component to inflammatory bowel disease, but that it does not show clear Mendelian inheritance ratios suggests any of the following, EXCEPT: A. the gene for the disease has incomplete penetrance. B. the gene for the disease has limited expressivity. C. the disease is polygenic. D. the gene for the disease is recessive.
D. the gene for the disease is recessive. (A) Incomplete penetrance occurs when an individual with a particular genotype does not express the expected phenotype. (C) Polygenic disorder is a genetic disorder that is caused by the combined action of more than one gene.
What are the direct and tropic hormones released from anterior pituatary?
FLAT PEG Follicle-stimulating hormone (FSH) Leutinzing hormone (LH) Adrenocorticotropic hormone (ACTH) Thyroid-stimulating hormone (TSH) --tropic hormones Prolactin Endorphins Growth Hormone ---direct hormones Most peptide and AA derivative hormones will have endings in -in (insulin, vasopressin, thyroxine T4, thyroxine) Most steroid hormones have names ending in -one (cortisol, corticosterone, testosterone, aldosterone).
What contributes to septic shock?
Figure 1 and the passage information indicate that the main purpose of Treatment 2 is to block the stimulation and proliferation of macrophages and to block their ability to release cytokines. Blocking this step of the cascade stops the events leading to hypotension and shock. Binding of the antibody to the macrophage does not stimulate its development and cytokine release; binding of the antibody to the macrophage prevents these results. It is true that the blockage of cytokine release could inhibit T-cell production, but this is not the rationale underlying the use of the Treatment 2 antibody. Want to increase BP with septic shock. Thermoregulation--- Fever = dissipation of body heat (vasodilation of capillaries to bring heat to the skin to be dissipated.
What is the difference between gram + and gram - bacteria?
Gram-positive (PURPLE): -Thick layer of peptidoglycan outside of cell membrane (one cell membrane) --contributes to purple color **peptidoglycan provides rigidity/support for the cell, contains cross-linked chains of sugars and amino acids. Gram-negative (PINK): -Thin layer of peptidoglycan in cell wall -Inner and outer membrane (outer membrane is a lippolysaccharide) -Gram negative bacteria are resistant to antibiotics.
Hypothesis B Although increases in resistance to blood flow can quickly increase blood pressure, increased pressure should presently act to initiate an effective corrective reflex involving the kidneys. The increased pressure should cause the kidneys to increase their output of fluid, and this should bring the pressure back to normal despite the persistent elevation in vascular resistance. The nervous system is probably not involved in this reflex. Failure of this reflex function may cause systemic hypertension. Assuming Hypothesis B to be correct, which of the following endocrine disorders would cause hypertension that could NOT be rectified by physiologically normal kidneys? A. An excess of aldosterone B. An excess of glucagon C. A shortage of thyroxine D. A shortage of insulin
I got confused what this question was asking. Essentially, this question is saying: "Essentially what its asking is, which of the following would lead to hypertension in an individual with a normal kidney?" REMEMBER The kidney is mainly responsible for: -regulating BP -regulating blood osmolarity -acid-base balance -removal of nitrogenous wastes -water reabsorbed to maintain BP + hydration This means that you can immediately eliminate (B) and (D) because both of these are functions of the pancreas to maintain blood glucose levels (via release of insulin/glucagon). Thyroxine (T4) is a hormone released by the thyroid gland to maintain homeostasis and metabolism. The answer is (A) because extra aldosterone means more Na+ resorption in the distal collecting tubule and collecting duct and thus increases water resorption too (does NOT affect osmolarity). This causes blood volume increases = BP increase = hypertension.
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
III. the disease was infectious. The scientists had hypothesized that the microbe pathogen was the causative agent of the disease even BEFORE the blood was analyzed, and viruses cannot undergo in vitro cultivation because they are obligate parasites. The fact that the disease was infectious was sufficient to put forth that hypothesis. Option III is correct, the fact that the disease was infectious caused the hypothesis to be formulated. The correct choice is B, III only.
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.
If you want the antisense RNA to have a therapeutic effect, it would need to be expressed whenever the target mRNA that it's meant to block is to be expressed. Thus the target gene and the antisense would have to be regulated in a similar manner. i.e. same promoter Just because a gene is physically near another doesn't mean if one is expressed the other will be. Gene expression is based on a handful of regulatory processes. 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. As a result, C is correct. A, B, and D are incorrect because the key feature determining whether an antisense drug will work is the timing of the expression of the antisense gene. This is controlled by specific regulatory elements, not necessarily the location of the antisense gene relative to the target gene. Thus, C is the best answer. You can have the antisense gene expressed while the sense is not, but if you have the sense expressed while antisense is not, you aren't going to have therapeutic effects!
Where are otoliths located and what sensory information do they convey?
In the utricle and saccule of the semicircular canals on top of the macula. Linear acceleration, gravity, change in head angle. When head and macula rate tilted, gravity pulls otoliths to deform the gel cap and bend cilia.
Can viruses undergo in vitro cultivation and if not, why?
NO! Viruses cannot be cultured in vitro because they are OBLIGATE parasites!
Are peptide hormones fast or slow acting? Are steroid hormones fast or slow acting? What about AA derived hormones?
Peptide hormone have a fast onset and is short-acting (like Insulin/ glucagon). Steroid hormones are slower but longer acting (like T and E). Some AA are fast acting and short lives (like Epi and Norepi) while some are slow and long acting (like T4 and triiodothyronine T3).
Does the antigen present on the surface of the cell or are they on chromosomes, DNA, or RNA?
Surface of cell.
What is osmotic pressure?
The "sucking" pressure of fluid from interstitial tissues.
Which of the following hormones is(are) directly required for spermatogenesis? Luteinizing hormone (LH) Follicle-stimulating hormone (FSH) Inhibin Testosterone
The Leydig cells secrete testosterone, an important male hormone. Testosterone acts on the Sertoli cells to promote maturation of sperm. FSH acts directly on the Sertoli cells to promote and maintain spermatogenesis.
What is a sera?
The blood serum of an animal, used especially to provide immunity to a pathogen or toxin by inoculation or as a diagnostic agent.
What is adaptive radiation?
The diversification of a single species of organisms into forms filling different ecological niches, such as what happened with the finches on the Galapagos islands. 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. A single rapidly changing environment does not encourage adaptive radiation. Rather, as the environment changes, the species would be more likely to collectively evolve into a single new species adapted to the new environment.
What is the purpose of bile?
The gall bladder stores bile produced by the liver and secretes it into the small intestine as needed. Bile acts as an emulsifier, facilitating fat digestion. When the gall bladder is removed, a patient will have reduced ability to digest fats.
How does the endosymbiotic theory explain the mitochondrial membrane structure in prokaryotes?
The inner membrane of a mitochondrion is analogous to the plasma membrane of a prokaryote. The enzymes for oxidative phosphorylation are embedded in the inner membrane. The endosymbiotic theory suggests that mitochondria are descendents of prokaryotes that were engulfed by endocytosis into a vesicle lined with a membrane derived from the cell membrane of the eukaryote host. This is the outer membrane. The inner membrane is the plasma membrane of the endosymbiotic prokaryote, so answer choice D is 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?
The larger pores in membrane 1 will allow heavier molecules to come through, however, as a whole, the filtration rate will decrease as the molecular weight increases. Smaller things will filter faster than bigger....filtration rate decreases as size increases. The passage says: "The dialysis membrane is made of a thin semiporous material, which allows diffusion of solutes with molecular weights up to 1000-2000 daltons, depending on the size of the membrane pores."
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.
The passage states: "After conjugation, the four sexually produced progeny divide by binary fission. Each daughter cell receives an exact copy of the micronuclear DNA and an uneven (but approximately equal) amount of DNA from the amitotic macronucleus. To minimize fluctuations in DNA content, small macronuclei undergo an additional S phase before division, and large macronuclei eliminate an S phase." When a vertebrate oocyte divides in meiosis the cytoplasm is distributed to the two daughter cells. Most of it goes to the daughter cell destined to be the ovum. The other daughter cells, the polar bodies, are cast off with little cytoplasm. The amount of nuclear material in each, however, is the same, so choice B is incorrect. The distribution of cytoplasm is uneven, as is the distribution of macronuclear DNA during the amitotic cell division in Tetrahymena. The answer is A, that the common feature is that both cytoplasmic division in ova and division of nuclear material in Tetrahymena are both uneven. Answer C is incorrect, Tetrahymena is unusual in having an amitotic division that requires an extra S phase; vertebrate ova have no such phase. Answer D is incorrect because there is no mitosis in the division of the macronucleus.
What is the function of the Na+/K+ pump?
The sodium-potassium pump sets the membrane potential of the neuron by keeping the concentrations of Na+ and K+ at constant disequilibrium. Basically, the membrane is more permeable to K+, which go out of leak channels, and there is a higher concentration of K+ in the neuron and more Na+ out of the neuron. K+ will leave through leak channels down the concentration gradient, but will be attracted to the negative membrane and will want to come back in, so it will be brought back in through the Na+/K+ pump (both ions moving against their concentration gradients).
The Ames test is often used in the initial screening of suspected carcinogenic compounds because it provides a good indication of the mutagenic characteristics of many chemicals. The test uses special strains of the bacterium Salmonella typhimurium that are nutritional mutants; they also lack a mechanism for DNA repair. When grown on a medium that lacks the amino acid histidine, the Salmonella test strains do not survive even though wild-type Salmonella grow well on this medium. During the Ames test, the suspected carcinogen is added to a histidine-deficient growth medium. If the chemical is a mutagen, some of the bacteria will back-mutate, and a visible colony will form. The usefulness of the Ames test can be improved when the growth medium contains rat-liver enzymes. The passage indicates that when Salmonella have back-mutated, they: A. contain a pigment that makes them visible. B. are capable of synthesizing histidine. C. will metabolize the carcinogen in the presence of light. D. lack histidine in their proteins.
The survival of the Salmonella on histidine-deficient media suggests that the strains have back-mutated and regained the ability to synthesize their own histidine due to exposure to the mutagen (B). The passage doesn't state anything about pigments (A) or the impact of light exposure (C)
What are the two ways that bacteria acquire antibiotic resistance?
There are two main ways that bacterial cells can acquire antibiotic resistance: 1. One is through mutations that occur in the DNA of the cell during replication. Mutations which make cells drug resistant are very rare, but the few drug-resistant bacteria that do develop such mutations flourish when the nonresistant cells are killed by antibiotics. 2. The other way that bacteria acquire resistance is through horizontal gene transfer: transformation, transposition, and conjugation. The genetic material that carries the resistance is the PLASMID.
Tissues that are exposed to the external environment all have what?
Tissues exposed to external environment all have mucosal membranes. All of the tissues listed contain an element of mucosa (e.g., the nasal cavity, vagina and anus).
True or false. Blood pressure is higher in the capillaries than the venules, veins, and vena cava.
True!
True or false. The parasympathetic nervous system and sympathetic nervous system release Ach from pre to post-ganglions but parasympathetic releases Ach from post-ganglions to target tissue while sympathetic nervous system released Epi/Norepi from post-ganglion to target tissue.
True.
True or false. Women tend to have lower bone densities compared to men.
True.
True or false. The left ventricle is thicker than the right and generates a greater pressure.
True. The difference in size between the thin wall of the right ventricle and the thick wall of the left ventricle is dramatic when viewed in cross section. The peak pressure in the left ventricle is 120 mmHg while the peak pressure in the right ventricle is only 25 mmHg. The right ventricle pumps blood through the lungs. The left ventricle pumps blood through the entire rest of the body. The difference in the size of the lung compared to the rest of the body suggests that the left ventricle develops more pressure and has a thicker wall than the right. The organs through which the left ventricle pumps blood are farther away from the heart than the lungs and resistance in a tube is inversely proportional to the length of the tube. This would also suggest a thicker wall for the left ventricle and greater pressure.
True or false. Prokaryotes have ribosomes.
True. Since they are not membrane-bound organelles, prokaryotes have them. Remember: Prokaryotes have 70-50s and 30s subunits. Eukaryotes have 80-60s and 40s subunits.
True or false. The closer the genome of two species, the more recently they diverged.
True. This is according to the molecular clock model.
What are polar bodies in oogenesis? Does spermatogenesis also have polar bodies?
When a vertebrate oocyte divides in meiosis the cytoplasm is distributed to the two daughter cells. Most of it goes to the daughter cell destined to be the ovum. The other daughter cells, the polar bodies, are cast off with little cytoplasm. The amount of nuclear material in each, however, is the same. Polar bodies allow resources to be focused on one egg to maximize quality over quantity. This makes sense for female germ cells which need to provide all the organelles for the egg cell to function and grow, while sperm cells only need to provide the paternal DNA. It's so that you can make one giant egg cell. The polar bodies are just the nucleus and a cell membrane basically, and all the organelles went to the actual egg cell.
Which germ layer gives rise to circulatory system?
mesoderm