Upgrade to remove ads
Kaplan Full-Length Test 2 Bio/Biochem
Terms in this set (41)
All of the following structures secrete digestive enzymes EXCEPT the:
a. oral cavity
d. small intestine
The esophagus is simply a conduit through which a food bolus passes from the pharynx to the stomach. There are no glands that secrete into the esophageal lumen.
A) oral cavity contains the salivary glands, which begin chemical digestion with the secretion of salivary amylase, or ptyalin. This enzyme hydrolyzes starch to maltose.
C) the pancreas secretes numerous digestive enzymes
D) The duodenum has mucosal glands called Branners glands, which secrete mucus to protect the small intestine from the acidity of gastric juices.
The rest of the small intestine contains mpits known as crypts of Lieberkuhn, which also have mucus-secreting glandular cells. In addition, intestinal glands secrete aminopeptidase and dipeptidases - enzymes that hydrolyze peptide bonds, and enterokinase, which converts trypsinogen to trypsin.
Which of the following organs is involved in neutralizing gastric acidity?
b. large intestine
Parietal cells: are in the stomach; secrete HCl
"Neutralizing secretions of the pancreas". The pancreas releases negatively charged bicarbonate ion into the duodenum. This neutralizes the incoming acid and protects the duodenal lining.
B) By the time the chyme reaches the large intestine, it is fully neutralized
Which of the following best explains why gastric pH must be precisely controlled?
a. gastric enzymes are most active at a low pH
b. high gastric pH stimulates the release of pancreatic secretions
c. low intracellular pH is necessary for proper parietal cell function
d. gastric juices create the optima environment for nutrient absorption in the large intestine
Gastric enzymes, such as pepsin, have optimum activity in an environment with a pH between 2 and 3. Enzymes are proteins; they rely on the appropriate ionic state of their primary amino acid structure for proper function. For example, if the substrate binding site of pepsin is altered by a change in electrical charge, then pepsin would not be able to hydrolyze those peptide bonds for which it is specific. The inherent nature of gastric enzymes makes them the most effective in an acidic environment. Extreme acidity, however, can cause denaturation of proteins, which is why there is that complex negative feedback system that maintains a narrow pH range within the stomach.
B) It is low pH, not high, that stimulates the release of pancreatic secretions
C) many proteins are denatured by acidity. A low intracellular intestine is neutralized by the bicarbonate ion in the duodenum; by the time it reaches the large intestine, the chyme is no longer acidic.
Secondly, nutrient absorption occurs in the small intestine, not the large intestine. The large intestine is involved in the absorption of salts and water.
Unlike nitric oxide release, release of serotonin from neurons involves the simultaneous release of between 1,000 and 10,000 molecules. Serotonin release illustrates which of the following transport mechanisms?
a. primary active transport
b. secondary active transport
c. facilitated diffusion
This question is testing your knowledge of how serotonin is released from neurons. The questions stem notes that it involves simultaneous release of a large number of serotonin molecules. That means the correct mechanism cannot involve transfer of one molecule at a time. Among the listed choices, only exocytosis--in which a vesicle fuses with the cell membrane and expels its contents--transfers large numbers of molecules.
A, B, C) primary active transport, secondary active transport, and facilitated diffusion, would move a small number of serotonin molecules (usually one to three) across the cell membrane
If the effects of NO occurred in arteries around the circulatory system, all of the following would occur EXCEPT:
"NO gas, which is toxic if inhaled, acts as a neurotransmitter and diffuses into nearby cells where it complexes with the heme subunit of soluble guanylate cyclase. This enzyme converts a molecule of GTP to cyclic-GMP. cGMP activates kinases that relax the smooth muscle of the corpus cavernosum, thus creating an erection"
a. cooling of the body
b. increase in blood pressure
c. decrease in venous blood volume
d. decrease in blood pressure and decrease in venous blood volume
NO causes relaxation of smooth muscle, and hence would cause expansion of arteries if present in the circulatory system. As arteries expand, blood travels out to the extremities. This results in heat loss through areas such as the hands and feet. The increase in arterial volume results in lower blood pressure.
A) dilation of blood vessels is one of the primary methods the body uses to get rid of excess heat. NO will indeed cause blood vessels to relax and dilate, and thus lose heat.
C) the blood is pulled out of venous circulation and into the now more voluminous arterial side so venous blood volume does decrease
Nitric oxide can enter cells by diffusing across the cell membrane. This is most likely because nitric oxide is:
a. relatively polar
b. relatively small
c. a free radical
d. concentrated inside the cell
This question wants to know why NO can diffuse across cell membranes. Diffusion generally requires that a molecules be either small or nonpolar, or preferably both. Nitric oxide, as a diatomic substance, is relatively small, and that's why it can diffuse across the cell membrane.
A) polar molecules generally have difficulty diffusing across cell membranes
D) Diffusion moves a molecules down its concentration gradient.
When Viagra inhibits phosphodiesterase-5, it:
a. binds irreversibly to the active site
b. reduces the Vmax of the enzyme
c. binds to an allosteric site
d. increases the Km of the enzyme
We know Viagra is a competitive inhibitor.
By definition, competitive inhibitors bind reversibly to the active site, which eliminates A and C.
Since reversible inhibition CAN be overcome by adding more substrate, the Vmax remains the same. The inhibitor actually increases the Michaelis-Menten constant, Km, the concentration at which the enzyme has half-maximal velocity.
At its isoelectric point, an amino acid in an electric field will:
a. migrate towards the anode
b. migrate towards the cathode
c. migrate towards the basic region
d. not migrate
At its isoelectric point, an amino acid exists as a zwitterion with positive and negative charges that cancel each other out. Because it has no net charge, it will not migrate in an electric field.
protein is negatively charged and driven toward the anode
Aged cheese and ripened tomatoes contain a higher concentration of MSG than their fresh counterparts. This is due to:
a. neural stimulation
b. T1R1+3 activiation
c. protein degradation
d. activation of the mGluR4 found in taste buds
The only way to increase the amount of free glutamate (like MSG) in a food source is through the breakdown of protein. As the cheese or tomatoes age, protein is degraded, releasing glutamate and therefore increasing ht perception of umami.
B) T1R1+3 enhances the taste of umami, but does not change the amount of MSG within a food source
The membrane receptor for glutamate opens an ion channel. Movement of ions through this channel is best described as:
b. facilitated diffusion
c. primary active transport
d. secondary active transport
Movement through a channel down the concentration gradient is an example of a passive process. Hence C and D are ruled out.
Passive transport: includes diffusion, facilitated diffusion, and osmosis.
Diffusion and osmosis: movement through the membrane of solute and solvent respectively.
Facilitated diffusion: requires a carrier molecule or pore. An ion channel is required for the transport of glutamate.
"MSG can also be isolated through fermentation of corn or sugar cane and is used as a food additive to enhance flavor"
Bacterial production of MSG from sugar molasses is:
a. aerobic and occurs after glycolysis
b. anaerobic and occurs after glycolysis
c. aerobic and occurs after the TCA cycle
d. anaerobic and occurs after the TCA cycle
Fermentation is an anaerobic process that occurs after glycolysis.
Question 21: The half-egg with a nucleus has:
a. the same number of chromosomes as the whole egg and twice the number of chromosomes as an autosomal cell
b. the same number of chromosomes as the whole egg and half the number of chromosomes as an autosomal cell
c.the same number of chromosomes as the half-egg without a nucleus and half the number of chromosomes as an autosomal cell
d. the same number of chromosomes as the half-egg without a nucleus and twice the number of chromosomes as an autosomal cell
An autosomal cell contains a full set of chromosomes; all cells except sex cells are autosomal. In other words, an autosomal cell is a diploid cell with 2 copies of every chromosome- one of maternal origin, one of paternal origin.
So how do the 3 egg types -- or gametes-- used in the experiment compare to an autosomal cell?
Well, most eukaryotic gametes, including those of sea urchins, are haploid cells. This means that each gamete has only one of each pair of chromosomes in its nucleus -- this is the result of meiosis; meiosis halves chromosome #, fertilization restores chromosome number, while mitosis maintains it. So the whole egg is a haploid cell, and therefore contains half the # of chromosomes as an autosomal cell. Likewise, the half-egg with a nucleus is also a haploid cell, since it also has a nucleus. The half-egg without a nucleus has no chromosomes.
In another experiment, the embryologist measured the rate of synthesis of a single protein and found that the synthesis rate varied with the amount of oxygen provided to the embryos. Which of the following enzymes is LEAST likely to affect the synthesis of the protein in question?
a. triose phosphate isomerase
b. cytochrome c oxidase
c. ATP synthase
d. pyruvate dehydrogenase
This question describes a modification to the experiment -- varying the amount of oxygen provided to the sea urchin cells and monitoring the amount of a specific protein being produced. Based on this information, the correct answer choice would be an enzyme that is necessary for the synthesis of a protein that isn't affected by the presence of oxygen.
Triose phosphate isomerase is involved in glycolysis. Glycolysis is a component of both aerobic and anaerobic respiration, meaning that the presence or absence of oxygen has no effect on glycolysis.
B, C, D) all enzymes that are a part of aerobic respiration an would result in increased protein synthesis rates in the presence of oxygen
A patient diagnosed with MCAD deficiency and no other health conditions is found to have a father that carries the same mutation, but a mother without evidence of mutation when a buccal swab is taken. Which of the following is the most likely reason?
"MCAD deficiency is an autosomal recessive disorder with variable expressivity and complete penetrance."
a paternal nondisjunction conferring 2 copies of the gene for MCAD
b. maternal germline mutation in the gene coding for MCAD
c. maternal nondisjunction deleting one copy of the gene coding for MCAD
d. multiple somatic mutations in the fetus of the gene coding for MCAD
A single event is almost always more likely than multiple events. Therefore the consideration should be isolated to events that would occur before division of the zygote. In this case, any loss of the fully functioning gene is likely to cause the symptoms. However, this still leaves maternal MCAD loss and maternal MCAD mutations as possibilities. To distinguish between them, recognize that the child does not have any other medical conditions. monosomy in any chromosome, as would result from maternal nondisjunction resulting in gene loss, is a serious medical condition.
Question 24: In a person with MCAD deficiency, if the body has excess glucose stores, which electron transfer intermediates are likely to accumulate? How does this change if the body is energy depleted?
a. NADH and FADH2 when high glucose, NAD+ and FAD+2 when depleted
b. NADH and FADH2 when high glucose, NADH and FADH2 when depleted
c. NAD+ and FAD+2 when high glucose, NAD+ and FAD+2 when depleted
d. NA+H and FAD+2 when high glucose, NADH and FADH2 when depleted
With excess glucose stores, the presence of a large proton gradient prevents the transfer of electrons in the ETC, thus causing a buildup of NADH and FADH2 while in a low energy state the need for ATP causes a decrease in these carriers and a corresponding increase in NAD+ and FAD+2.
Question 25: Why is MCAD deficiency more likely to appear earlier (as soon as a few days after birth) in breastfed infants? Initially, breastfed infants receive a:
a. larger supply of medium chain fatty acids than bottle fed infants
b. smaller supply of medium chain fatty acids than bottle fed infants
c. larger number of carbohydrate calories than bottle fed infants
d. smaller number of carbohydrate calories than bottle fed infants
The initial period of breastfeeding occurs before milk let-down in a woman. Therefore the infant is receiving a limited amount of calories from colostrum, mostly consisting of protein. In bottle-fed infants, there is immediate consumption of a higher calorie, higher carbohydrate diet. This diet is consistent with the MCAD deficiency treatment, therefore preventing symptoms in these infants. Additionally, a switch to fat metabolism would trigger the use of the MCAD enzyme, and a low carbohydrate load would facilitate that switch by activating glucagon and other hormones.
The amount of medium chain fatty acids should not contribute either way to MCAD deficiency symptoms because they will not be broken down. The only symptom that is expected to increase in the case of medium chain fatty acid excess is hepatomegaly as they accumulate.
Genetic testing for MCAD deficiency can be challenging because multiple, distant sites within the gene can be mutated to produce disease, while mutations at other sites within the gene are benign. What does this suggest about the necessary enzyme structure?
a. primary structure is most important for enzyme function
b. secondary structure is most important for enzyme function
c. tertiary structure is most important for enzyme function
d. quaternary structure is most important for enzyme function
From outside knowledge, most inactive enzymes have modifications of the active site, which is part of the tertiary structure.
A, B) there are mutations that are benign. These mutations would change primary structure (and possibly secondary structure), however, since the enzyme is still functional, the tertiary structure must be intact.
D) the passage states that MCAD is monomeric, thus quaternary structure is not a concern.
Researchers have bred a line of "knockout mice" where the function of a single gene has been made inoperative. They discover that the test subjects are missing 6 different proteins that are found in wild type mice. What might account for this difference?
a. the primary transcript that is normally derived from the "knockout gene" can be processed six different ways by nuclear spliceosomes.
b. the polypeptide that is derived from the "knockout" gene can take on 6 different tertiary structures affectin gits functionality
c. degeneracy within the genetic code allows a single codon to translate multiple amino acids yielding 6 different proteins from the "knockout" gene
d. multiple different start codons within the single "knockout" gene allow for the processing of 6 different primary transcripts
Transcription of a specific gene will always result in the same primary transcript (pre-mRNA). Hence, D cannot be the right answer. However, the primary transcript can be processed in several different ways by nuclear spliceosomes, leading to the formation of differing mRNAs, and subsequently, differing proteins.
One species benefits and the other neither benefits nor suffers.
After an infection, such as by C. difficile, the normal gut flora "re-colonize" the digestive tract. Those re-colonizing bacteria are most likely harbored by the:
a. terminal ileum
Normal gut flora "hang out" in the cecum, the blind outpouching of the large intestine
A) the terminal ileum is part of the small intestine
C) the jejunum is the middle portion of the small intestine, which is primary involve din absorption
D) the rectum is the site of solid waste storage, not a harbor for bacteria
Duodenum - complete the first phase of digestion
Based on the information in the passage, which of the following is NOT true of Bacteroides?
a. Bacteroides cells are facultive anaerobes
b. Bacteroides is a Gram-negative bacteria
c. Samples of Bacteroides would be difficult to distinguish from Prevotella on the basis of light microscopy
d. The cell wall of Bacteroides is relatively thin, and contains a periplasmic space
This question is looking for something not true of Bacteroides. The 3 wrong answers, then, will be true. It's difficult to make an exact prediction here, so cycle through the answer choices.
The passage tells us that the ratio of obligate anaerobes to facultative anaerobes is more than 100 to 1, and it also tells us that Bacteroides accounts for about 30% of gut bacteria. These 2 statements taken together mean that Bacteroides is an obligate anaerobe.
B) Table 1 states that Bacteroides is Gram-negative
D) definition of a Gram-negative organism: cell wall of Bacteroides is relatively thin, and contains a periplasmic space
C) states that Bacteroides and Prevotella are Gram-negative bacilli
Gram-negative vs Gram-positive
Difference in the structure of their bacterial cell wall. Gram-positive bacteria do not have an outer cell membrane found in Gram-negative bacteria.
Gram-positive: cell wall is high in peptidoglycan which is responsible for retaining the crystal violet dye.
Gram-negative: all bacteria have an inner cell membrane, gram-negative bacteria have a unique outer membrane. Generally more resistant to antibiotics than gram-positive bacteria -- excludes certain drugs and antibiotics from penetrating the cell.
Which of the following activities would you expect to increase in a tumorigenic cell?
I. mRNA synthesis
II. Ribosomal assembly
III. Cell division
Answer: I, II, III
According to the passage, eukaryotic cells contain genes known as proto-oncogenes, which normally code for proteins involved in the regulation of growth. If any of these proto-oncogenes become transformed into an oncogene, then that cell is said to be tumorigenic.
A tumorigenic cell is one that gives rise to a tumor, and unlike normal cells, tumor cells don't obey the rules of normal cell growth and divide indefinitely. Thus, in a tumorigenic cell, you would expect to see an increase in all of the activities and processes associated with cell growth and division, such as mRNA synthesis (transcription) and ribosomal assembly. Likewise, cell division would also increase, since tumor cells replicate at an accelerated rate.
Based on the information in the passage, cellular proto-oncogenes can become tumorigenic oncogenes by all o the following mechanisms EXCEPT:
a. a mutation that results in the synthesis of a faulty protein
b. a chromosomal translocation that produces an excess of a protein
c. binding of complementary nucleic acid sequences to proto-oncogene transcripts
d. a mutation that causes gene amplification of the proto-oncogene
In order for a proto-oncogene to transform a cell, oncoproteins must be produced from proto-oncogene transcripts. But if there are complementary nucleic acid sequences present, the proto-oncogene mRNA will base pair with the nucleic acids before it can be translated and produce oncoproteins.
A) Cellular proto-oncogenes (according to the passage) often become tumorigenic via a point mutation that leads to the formation of a defective protein. This is a true statement.
B) Paragraph 2 states that a mutation that causes a proto-oncogene to produce an excess of its protein product, such as a chromosomal translocation, will convert the proto-oncogene into a tumorigenic oncogene. This is a true statement
D) Paragraph 3 states that a mutation that causes the proto-oncogene to undergo gene amplification would also cause an excess of protein product and convert the proto-oncogene into a tumorigenic oncogene. This is a true statement.
During replication repair, enzymes can reverse an error that would lead to the formation of an oncogene from a proto-oncogene. Which enzyme is responsible for the last step in the DNA repair?
b. DNA ligase
d. DNA polymerase
During base repair, a base is excised by a restriction enzyme, replaced by a polymerase, and finally linked o the bases around it by DNA ligase.
A) Primase is a part of replication, but it is not involved in the repair process
C) topoisomerase relaxes and winds/unwinds DNA for synthesis. it is not directly involved in DNA repair
Before being sent to the cell membrane, the EDAR protein is processed in:
a. the rough endoplasmic reticulum
b. the smooth endoplasmic reticulum
c. the Golgi complex
d. the lysosome
Proteins are processed for transport to the cell membrane in the Golgi complex.
A) the rough ER is the site of much protein synthesis, but not the processing that precedes transport to the cell membrane
B) the smooth ER is not involved in protein synthesis or trafficking
D) the lysosome digests proteins, but does not process them for transport
Question 40: The protein that hair is primarily composed of is:
Since hair is a specialized form of epidermis, it, too, consists of dead cells. After the cell dies, what's left is primarily keratin.
B) Collagen is the most prevalent protein in the human body as a whole, but not in hair.
C) chitin is not produced by humans; it is found in the call walls of fungi and the exoskeletons of arthropods
D) fibrin is a protein involved in the clotting of blood
Missense mutation versus silent mutation
Silent mutation -- results in the same amino acid
Missense mutation -- results in a different amino acid
Question 42: A scientist studying the EDAR gene finds that the wild type mRNA for position 370 is GUA. Assuming that only point mutations without a frameshift occur, what is the minimum number of mutations needed to introduce a stop codon at this position?
d. an abnormal stop codon can only be introduced by frameshift mutations
This question is asking for minimum number of mutations needed to convert GUA to a stop codon. The 3 stop codon are UGA, UAG, UAA.
Question 43: A mountain climber living at sea level ascends to a very high altitude during the course of a day long climb. By the end of the day, all of the following acclimatizations will occur EXCEPT:
a. decreased heart rate
b. inhaling more air than normal
c. increased tidal volume
d. increase secretion of erythropoietin
a decrease in heart rate would decrease the perfusion of tissues with blood, thereby decreasing oxygen delivery. This would not happen to acclimatize the climber's body to the increased altitude.
B) Inhaling more air than normal will compensate or the fact that the inhaled air has less oxygen than normal
C) the body will make rapid adjustments to the low pO2 of high altitude in order to maintain the delivery of oxygen to the tissues. Tidal volume will increase in order to increase the minute volume (amount of air inhaled and exhaled in a minute)
D) the body will secrete erythropoietin to increase the number of erythrocytes in the blood and thus increase the oxygen carrying capacity of the blood
Question 44: A scientist is planning to study the effects of osmotic pressure on membrane transport processes. The scientist wants to create a solution of NaCl that exerts 1 atm of osmotic pressure at 25C. What concentration of NaCl is needed to accomplish this? [Note: the ideal gas constant R = 0.0821 L x atm x mol^-1 x K-1]
a. 0.02 M
b. 0.04 M
c. 0.08 M
d. 56 M
The formula for osmotic pressure is π = iMRT, where i is the van't Hoff factor (equal to the number of particles in solution per original formula unit), M is the molarity of the solution, R is the ideal gas constant (0.0821 L x atm x mol^-1 x K^-1), and T is the temperature in kelvins.
In this case, the question stem tells us that T = 298 K, which we can round off to 300, and i = 2 (since NaCl dissociates into Na+ and Cl- ions). Thus, since we want a final pressure of 1 atm:
1 atm = 2M (0.0821)(300)
1 = 2 x 0.0821 x 300 M
so 1 is approximately equal to 50 M and M = 0.02. Thus we need a solution of about 0.02 M NaCl to generate a pressure of 1 atm. This matches choice A.
B) omits the value of i = 2 in the formula for osmotic pressure. It would be correct for a solute that does not dissociate, such as glucose.
D) this is solving the wrong problem: instead of finding the concentration corresponding to 1atm of osmotic pressure, it gives the osmotic pressure corresponding to 1 M solution.
Which of the following lists a pair of physiological changes that would both be likely to increase the rate of pyruvate formation in glycolysis?
a. an increase in fructose 6-phosphate levels and an increase in glyceraldehyde 3-phosphate
b. an increase in glucose 6-phosphate levels and an increase in acetyl coenzyme A levels
c. a decrease in ATP levels and an increase in citrate levels
d. a decrease in ATP levels and a decrease in glucose 6-phosphate levels
This question is asking for a pair of physiological changes that would increase levels of pyruvate formation in glycolysis. As a general rule, an increase in substrates that precede pyruvate in glycolysis, or a decrease in substrates that follow it. The only choice in which both changes agree with one of these two is choice A: Both F6P and G3P are substrates that precede pyruvate in glycolysis.
B) an increase in acetyl CoA should create a "bottleneck", decreasing pyruvate formation
C) an increase in citrate levels should decrease pyruvate formation
D) a decrease in G6P depletes a prior step in glycolysis, and thus should decrease formation of pyruvate
An 18-carbon fatty acid undergoes beta-oxidation as shown below:
[C18 acid]-S-CoA + FAD + NAD+ + H2O + CoA-SH --> [C16 acid]-S-CoA + CH3CO-S-CoA
An experimenter adds a radioactive label to C-1 of the 18-carbon acid and to the sulfur atom in the coenzyme A used as a reactant. Where will those labels likely appear in the products?
a. the labeled carbon will appear in the C16 acid while the labeled sulfur will appear in acetyl coenzyme A
b. the labeled carbon will appear in acetyl coenzyme A while the labeled sulfur will appear in the C16 acid
c. both labels will appear in the C16 acid
d. both labels will appear in the acetyl coenzyme A
This question is testing your knowledge of fatty acid oxidation, which is known as Beta-oxidation because it cleaves the carbon-carbon bond before the beta carbon of the acid. (Remember that because of the peculiarities of alpha, beta notation, the beta carbon is C-3, not C-2!). The question wants us to trace the fate of 2 labels, a carbon atom and a sulfur atom. The carbon atom is C-1 of the acid; as noted above, this will end up in the acetyl coenzyme A. The sulfur in the labeled coenzyme A will end up attached to the C16 product.
C-1 is the carboxylic acid carbon, not the carbon at the end of the hydrocarbon tail. The sulfur label should thus end up attached to the remainder of the carbon chain.
Question 47: The binding of oxygen to hemoglobin does not follow the rules of Michaelis-Menten kinetics, primarily because:
a. hemoglobin does not modify oxygen
b. hemoglobin can bind more tightly to carbon monoxide
c. hemoglobin exhibits allosteric effects
d. oxygen is a gas at standard temperature and pressure
The key assumption on which Michaelis-Menten kinetics are based is that there is one active site capable of binding substrate. Hemoglobin, however, is an allosteric protein with 4 subunits, and exhibits cooperativity: when one oxygen binds, the other sites bind oxygen more easily. While all of the answer choices are true statements, C is the only statement that explains why Michaelis-Menten kinetics don't apply to the binding of oxygen to Hb.
D) Michaelis-Menten kinetics applies to dissolved gases as well as to any other aqueous solute.
The open-chain structures of D-glucose and D-fructose are shown below. D-glucose and D-fructose are most precisely characterized as:
d. constitutional isomers
Remember the classification scheme for isomers. The first question to ask when comparing 2 structures is "Do they have the same connectivity?" In other words, are the bonds exactly the same? In this case, the answer is no--the carbonyl is terminal in glucose, but internal in fructose. Therefore, glucose and fructose must be structural isomers, also known as constitutional isomers--which is choice D.
A) anomers are different forms of the same carbohydrate. (alpha-glucose and beta-glucose)
B) to be diastereomers, they must be optical isomers that are identical at one or more chiral centers and different at one or more chiral centers, which is not the case for fructose and glucose
C) while D-glucose and D-fructose are in the same D/L family, they do not differ at only one carbon, as epimers do
Enantiomers: mirror image
Diastereomer: differ at some stereocenters but not at others so are not mirror images.
Excessive blood glucose levels lead to polyuria, an increase in urine output. A possible explanation for this is that the presence of glucose in the renal tubule directly causes a decrease in:
a. the number of aquaporins closing in the collecting duct
b. reabsorption of Na+ in the distal convoluted tubule
c. movement of water through cells in the ascending limb of the loop of Henle
d. movement of water through cells in the descending limb of the loop of Henle
One of the classic symptoms of type 2 diabetes is polyuria, excess water excretion. But how does this happen? Higher than normal amounts of glucose in the urine means that the urine osmolarity will be higher than usual.
This will result in less reabsorption of water from the filtrate in the nephrons. C and D are the only statements that are consistent with this reasoning. However, the ascending limb is almost impermeable to water, which means choice D is the correct answer.
A) an increase in open aquaporins would increase reabsorption of water from urine, leading to a decrease in urine output, which is the opposite of polyuria.
B) glucose does not directly affect Na+ reabsorption, aldosterone does.
C) the ascending limb of the loop of Henle is impermeable to water.
Question 52: Type 1 diabetes typically occurs as a result of autoimmune destruction of:
a. acinar cells
b. alpha cells in the islet of Langerhans
c. beta cells in the islet of Langerhans
d. chief cells
What cells are destroyed in type 1 diabetes? Type 1 diabetes involves destruction of cells that produce insulin. Those are the beta cells of the islets of Langerhans of the pancreas.
Type 1: autoimmune disease that destroys the beta cells of the pancreas so insulin is not produced.
Type 2: insulin resistance. Beta cells must produce more insulin, and they are overworked and will shut down
Question 53: In a patient with failed closure of the rostral neural tube, which of the following is most likely to be unaffected?
a. lower extremity deep tendon reflexes
b. ability to discriminate between tastes
c. upper extremity pain sensation
d. vestibular sense
As described in the passage, a failure to close the rostra neural tube results in anencephaly. As can be determined from the name of the condition and the description of the developmental anomaly, the brain fails to form. Therefore, functions controlled exclusively by the spinal cord will remain intact. This includes certain pain and monosynaptic reflexes, like the deep tendon reflex.
Sensation is a cortical function, rather than a spinal function, so choice C is incorrect.
Question 54: All of the following proteins are likely to be exceptionally active in and around the cells of the primitive neural tube EXCEPT:
a. the sonic hedgehog protein
b. cell adhesion molecules
c. cytoplasmic motors
As mentioned in the second paragraph, during neural tube development, sonic hedgehog proteins play an important role and would be expected in exceptionally high amounts. In the 3rd paragraph it states that during closure, there is "migration and replication of cells" which would imply a need for cell adhesion molecules and motors. Immunoglobulins are not produced by the neurula.
Question 55: The type of signaling described in paragraph 2 is:
A signaling cascade in development which changes the structure or function of developing tissues through chemical mediators is called induction.
Transduction: process by which foreign DNA is introduced into a cell by a virus or viral vector. An example is the viral transfer of DNA from one bacterium to another
Question 56: At what point in the developmental scheme shown in Figure 1 do the cells lose totipotency?
a. before phase 1
b. between phases 1 and 2
c. between phases 3 and 4
d. after phase 4
Totipotency: Ability of a cell to develop into any mature cell type. It is generally retained only in the very early embryo. The scheme in the figure has already designated tissue types into specific categories like neural plate, surface ectoderm, and notocord. Therefore, these cells are no longer totipotent. They do remain pluripotent until about phase 4, but this is not the question that is being asked.
Pluripotent: capable of giving rise to several different cell types
Assuming classical Mendelian inheritance, how can one differentiate between a homozygous dominant individual and one who is heterozygous for the dominant trait?
a. by crossing each individual with a known homozygous dominant and examining the offspring
b. by crossing each individual with a known homozygous recessive and examining the offspring
c. by crossing each individual with a known heterozygote and examining the offspring
d. both B and C
to differentiate between a homozygous dominant and a heterozygous dominant for a trait that exhibits classic dominant/recessive Mendelian inheritance, one must perform a cross that results in offspring that reveal the unknown parental genotype; this is known as a test cross.
If we cross the homozygous dominant with a homozygous recessive, we will get 100% phenotypically dominant offspring; if we cross the heterozygous dominant with the homozygous recessive, we will get 50% phenotypically dominant and 50% phenotypically recessive offspring.
Thus using a homozygous recessive as a test crosser will allow us to distinguish between the two.
We can also use a known heterozygote as the test crosser because when this is crossed with the homozygous dominant, 100% phenotypically dominant offspring are produced, and when it is crossed with the heterozygote, the phenotypic ratio of the offspring is 3:1 dominant: recessive.
THIS SET IS OFTEN IN FOLDERS WITH...
Kaplan Practice Test 2 Chem/Physics
Kaplan Full-Length Test 1 Physics
Kaplan Full-Length Test 3 Physics/Chem
Kaplan FT Test #2
YOU MIGHT ALSO LIKE...
AAMC Biological and Biochemical Foundati…
AAMC FL 1 Bio
AAMC Full length 2 - Biochem Review
biology/biochemistry practice exam 4
OTHER SETS BY THIS CREATOR
Genetics - Lecture 4
MCAT quiz questions
Emily Vietnamese Words #3
Emily Vietnamese Words #2
OTHER QUIZLET SETS
Atividade Literatura - As Gerações Românticas
Econ test 12/8/20 stocks
EMT FINAL EXAM REVIEW
Chapter 22 exam