MCAT Chem/Phys
Which of the following is closest to the bond angle between the carbons in a molecule of acetone? A. 90º B. 109.5º C. 120º D. 180º
Acetone has three substiuents and no lone pair, therefore it would be sp2 which is 120 degrees.
It is found that small to moderate doses of I-131 are more likely to cause thyroid cancer than extremely large doses. Which of the following provides the best explanation for this observation? A. Small doses of I-131 are not absorbed by the body. B. Extremely large doses of I-131 are not absorbed by the body. C. The amount of radiation released by large doses of I-131 is enough to damage DNA without killing cells. D. The amount of radiation released by small doses of I-131 is enough to damage DNA without killing cells.
Answer was in the passage. If it takes large amounts to cause cancer and kill people, then it would take small dosages to damage DNA without killing the cells. Therefore, D is the correct answer
Why did the researchers choose to study pediatric, rather than adult, thyroid cancer cases? A. Thyroid cancer is not otherwise present in children. B. Thyroid cancer is not otherwise present in adults. C. Children receive a higher relative dose of I-131 at the same contamination level. D. Children receive a lower relative dose of I-131 at the same contamination level.
Because children weigh less than adults, they would receive a higher concentration. Think of when (not really) a kid would drink alcohol they would get drunk quicker because they weigh less. Therefore, C. is the correct answer
What is the chemical formula of gypsum? A. CaSO3 B. CaSO4 C. Ca2SO3 D. Ca2SO4
Calcium sulfate (given in the passage). Think charges. Ca has a charge of 2+ and sulfate (SO4) has a charge of 2-. Therefore, B. CaSO4 is the correct.
What is the approximate molarity of NaCl in ocean water, if the density of ocean water is 1.028 kg/L? A. 0.026 M B. 0.62 M C. 0.96 M D. 9.6 M
Change kg/L to g so move the decimal three places. Then look for key words "ocean water" so then look for numbers (3.5% -->0.035) to get g of NaCl. Then divide grams by grams (mm of NaCl). Round and round back up to get the final answer. B. 0.62 M
Which of the carbon(s) in 2-methylundecanal below is (are) chiral? A. 1 only B. 2 only C. 1 and 2 D. 2 and 3
Chirality- 4 different constitutes to a carbon. Therefore, B is the correct.
Lathyrism is known to target and degrade collagen in the lower limbs. Which pair of amino acids would best provide disulfide links to stabilize the folded form of collagen? A. C and C B. M and M C. M and C D. S and S
Cysteine is the only amino acid that forms a difsulfide bridge. Therefore, A is the answer.
Vinblastine is a microtubule-disrupting drug that inhibits tubulin polymerization. Which of the following processes would be directly inhibited upon vinblastine treatment? I. Phagosome transport to the lysosome II. Mitosis III. Meiosis IV. Electron transport A. II only B. I and IV only C. II and III only D. I, II, and III only
Definition of microtubules. Microtubules transport vesicles and position organelles in the cell. They can help transport phagosomes to the lysosome so I is right. This takes away A and C. They also form the spindle apparatus in mitosis and meiosis. Therefore, the answer is D.
Boiling chips and vacuum distillation, respectively, are used in distillations to: A. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled. B. lower the boiling points of the substances to be distilled; work synergistically with the vacuum system to further lower the boiling points. C. lower the boiling points of the substances to be distilled; provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating. D. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; speed up the distillation process by vacuuming the first distillate out of the apparatus.
Definitions, but the passage also gives you a quick definition of vacuum distillation; that it lowers the boiling. So that takes away B, C and D. That would only leave A as the correct answer.
Consider the net reaction leading to formation of fructose 6-phosphate from glucose and ATP. ΔG° for the reaction equals: A. +3.96 kcal/mol, and Keq for the reaction is greater than 1. B. -3.60 kcal/mol, and Keq for the reaction is greater than 1. C. -3.60 kcal/mol, and Keq for the reaction is less than 1. Show Explanation D. +3.96 kcal/mol, and Keq for the reaction is less than 1.
Delta G was less than zero so therefore the answer has to be negative. It required prior knowledge to know that the ATP between these two reactions was -3,600. Therefore, the answer is B
The meal that the scientists provided to each of the volunteers at t = 10 hours contained five eggs weighing 50 g each. Approximately how much methionine was served to each volunteer? A. 10 mg B. 1 x 103 mg C. 3 x 103 mg D. 6.25 x 104 mg
Didn't have time to read the last few passages so most was guessing. The passage states that 400 mg of met per 100 g of weight. 5 eggs at 50 g weighs about 250 g total. 250g x 400/100)=1000=1x10^3. Therefore, the answer is B.
Based on Figure 1, adding salt to water causes the boiling point of the water to: A. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. B. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure that is greater than the external pressure. C. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. D. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure that is less than the external pressure.
Figure 1 shows a phase diagram. The diagram shows that vapor increases with pressure. The second part of the question is the definition of boiling points. So, A would be the answer.
The transcription factor AP-1 is a heterodimer consisting of c-jun and c-fos. C-jun and c-fos are soluble proteins that can be localized to either the cytosol or nucleus of a cell. C-jun and c-fos dimerize through a leucine zipper motif. In a leucine zipper motif, every 7 amino acid residues, or 2 full turns of an alpha helix, are leucine resides. Leucine and other amino acids on one face of the helix come together to form an opposite alpha helix that has a similar arrangement of leucine and other amino acids. Which solvent would be LEAST favorable for c-fos/c-jun dimerization? A. Hexane B. Ethanol C. Water D. Phosphate-buffered saline
For long questions, pick out what they are really asking. They are asking for which solvent that would react favorably with leucine. Since leucine is nonpolar, it would also like something nonpolar. Hexane is the only nonpolar substance listed. Therefore, A is the answer.
What are the formal charges on the atoms in the molecule below? (Note: the molecule has not been labeled with overall formal charge and may have an overall formal charge that is 0 or a value other than 0.) A. I: +1; II: +1; III: -1 B. I: -1; II: +1; III: +1 C. I: +1; II: 0; III: -1 D. I: +1; II: -1; III: -1
Formal charge equation= # of valence electrons-sticks-dots. That would be the roman numeral on top of the periodic table. Therefore, the answer would be A.
What effect will the addition of a fluorine substituent have on carbocation stability? A. The fluorine group will be destabilizing because it is highly electronegative. B. The fluorine group will be stabilizing because it is highly electronegative. C. The fluorine group will be destabilizing because it has additional lone pair electrons. D. The fluorine group will be stabilizing because it has additional lone pair electrons.
From the passage, we see that fluorine is destablizing because it has a + RSE of 2.9. This gets ride of B and D. We know that all elements want to be like noble gases, so therefore D is eliminated. The only answer left is A.
If a person drank a large quantity of hypersaline ocean water, the person could die because absorption of salt into the blood will cause it to become: A. . hypotonic compared with the cytosol of the body's cells, causing osmosis of water into the cells. B. hypertonic compared with the cytosol of the body's cells, causing osmosis of water into the cells. C. hypotonic compared with the cytosol of the body's cells, causing osmosis of water out of the cells. D. hypertonic compared with the cytosol of the body's cells, causing osmosis of water out of the cells.
Hypertonic- cells would shrink, water would go out of the blood into the cell Hypotonic- cells would swell and could burst and the environment would cause the water to go into the cells. Water would follow NaCl into or out of the cell. D. cells would be hypertonic compared to the environment, which would cause osmosis of water.
The researchers originally wanted to recruit additional volunteers with sulfur-deficient diets for this study. What is the most likely reason this request was refused? A. It was difficult to find patients with sulfur-poor diets in the area where the study was conducted. B. The researchers found many willing participants who had sulfur-rich diets, and it therefore seemed appropriate for that to be the focus. C. The mechanism of how L. sativus affects sulfur-deficient patients is already clear. D. The health risks of providing L. sativus to sulfur-deficient patients were too high.
I didn't have enough time to read this passage, but the passage clearly states answer D, which is the correct answer.
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is 0.25. When the block is halfway down the ramp, the child pushes down on the block perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the block from starting to slide down the ramp? A. mg sin θ B. 0.25 mg cos θ + mg sin θ C. mg sin θ - 0.25 mg cos θ D. [(mg sin θ) / 0.25] - mg cos θ
I got this one right because I guessed. Visualize the problem. Create a phase diagram and see what is being asked while checking angle measures. Therefore, the correct answer is D.
How many moles of captopril were present in the original analyte solution tested? A. 7.5 × 10-5 moles B. 1.5 × 10-4 moles C. 7.5 × 10-3 moles D. 1.5 × 10-2 moles
I have no clue how I got this right, it was a complete guess. Think of titration curve and the equation for it (M(base) x V (base) = M(acid) x V (acid)= moles of acid. The equivalence point of captopril is given but it is not a monoprotic acid (acids that can release only one proton per molecule and have one equivalence point). First plug in numbers given in the problem into the equation and then use that number to find moles. Therefore, the answer is B.
If the students perform an enzyme inhibition assay using captopril, which of the following changes in the kinetic parameters of ACE should be expected? A. Vmax decreased; Km unchanged B. Vmax decreased; Km increased C. Vmax unchanged; Km decreased D. Vmax unchanged; Km increased
I picked A; but I didn't look into the passage to find that it literally says "captopril is a 'competitive inhibitor of ACE'". So, this would come down to remembering the Vmax equations and if Km would decrease or increase. Therefore, the answer is D because in competitive inhibitors Vmax is unchanged and Km is increase
Which of the following is most likely to result in ankle injury? A. 700-N normal force exerted by a flat horizontal surface on the feet B. An anterior force of 450 N at the pivot point of the ankle C. A 900-N normal force exerted by a flat horizontal D. An anterior force of 400 N at a distance of 10 cm from the pivot point of the ankle
I picked C because I was thinking that a person was running... but that was just an assumption because it doesnt mention that the person is walking or running. Therefore, D is the correct answer because of you pivot or move that would require torque which would show that you are in motion.
Which of the following correctly describes the orbital hybridization of XeF4 and NH3, respectively? A. sp3d2, sp3 B. sp3, sp3 C. sp3, sp2 D. sp3d2, sp2
In orbital hybridization, count the number of bonds and lone pairs around the central atom. Nitrogen has 4 total regions and therefore 4 orbitals to hybridize (s, p, p, p). So this gets rid of C and D. For XeF4, it has 4 bonds and two lone pairs because its valence is 36. therefore it has (s, p, p, p, d, d). Therefore, the correct answer is A.
The preferred ion configuration of many elements on the periodic table is determined by: A. the electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same core electron configuration as this noble gas. B. the electronegativity of the element directly above it, within the same group; elements will gain or lose electrons until they have an electronegativity that matches the period immediately above them. C. the electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same valence electron configuration as this noble gas. D. the relative stability of the nearest d subshell; elements will gain or lose electrons until the outermost d subshell available is stable.
Ion configuration would be the electron configuration of the noble gases and they need to gain or lose electron to have their valence electrons. Therefore, C is the correct answer
A mass of 10 kg is dropped from a height of 20 m. Ignoring air resistance, what is the maximum speed achieved by the mass? (assume g = 10 m/s2) A. 10 m/s B. 20 m/s C. 200 m/s D. 400 m/s
KNOW EQUATIONS. when something is in the air it has potential energy. The only time it has kinetic energy is right before it hits the ground. Therefore both equations are needed. PE=mgh= 10 kg x 10 m/s2 x 20 m= 2000J KE= 1/2mv2 2000J= 1/2(10 kg) xv2 4000= 10v2 400=v2 20=v
The association constant, Ka, of Epithelial Growth Factor Receptor (EGFR) binding to Epithelial Growth Factor (EGF) is 5.61 x 106. What is the Keq of EGFR + EGF → EGFR-EGF? A. 5.61 x 106 B. 1.78 x 10-7 C. 1.12 x 107 D. 8.42 x 10-6
Ka= Keq so therefore it would be the exact same number. Hence, A is the correct answer.
The reaction to produce the "X" form of compounds 1-4 from standard-state elements will be spontaneous under what conditions? A. The reactions will be spontaneous under all conditions. B. The reactions will be spontaneous at high temperatures. C. The reactions will be spontaneous at low temperatures. D. The reactions will not be spontaneous under any conditions.
Look at the passage for correct answers. We would have to determine the Gibbs Free energy and we see that it favors + delta G and the nonspontaneous reaction. therefore the answer is C.
During the force test with the mechanical stirrup, at what distance from the ankle joint was the force applied? A. 57 mm B. 60 mm C. 110 mm D. 133 mm
Look in the passage for some of the units. The equation for torque is T=rFsin(theta). sin (90)=1. Then, solve for r which would be 13/98=0.133 which is D.
A book rests horizontally on a table. The book experiences a gravitational force of mg due to the earth's gravity. According to Newton's third law: A. the book experiences a normal force of mg pushing up due to the table. Show Explanation B. the earth experiences a gravitational force of mg from the book. C. the table exerts a gravitational force of mg on the earth. D. the earth exerts a normal force up on the table equal to mg plus the weight of the table.
Newton's third law says that Fa on b= -Fb on A. The force would be the Earth pulling on the book and is opposite of the book pulling on the earth. Therefore, B would be the right answer.
Nitrogen primarily exists in the atmosphere as a diatomic gas. Which of the following is true about this form of nitrogen? A. The presence of a lone pair of electrons on each nitrogen atom in the molecule allows it to act as a strong Lewis base. B. The triple bond of electrons creates a region of high electron density that allows N2 to be very reactive as a nucleophile. C. Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions. D. Atmospheric nitrogen reacts spontaneously with carbon dioxide, which keeps atmospheric CO2 levels at a relatively low 0.04% (on a molar basis) of the atmosphere.
Nitrogen is tested heavily on the exam. Be familiar with N2 and that it is very inert and nonreactive; therefore can be used in my experiments. Therefore, the answer is C.
A circuit is constructed with a 12-V battery and four identical resistors, each with a resistance of 16 Ω, hooked up in parallel. What is the total power dissipated by the circuit? A. 4 W B. 4 J C. 18 J D. 36 W
P=W, therefore, immediately B and C can be eliminated because they are in J which is energy. P=V2/R. Each I would be a fraction, so 1/16 x4. =1/4 and flip it to be 4. 12^2/4=144/4=36W. Therefore, the answer is D.
What is the lift force for the vesicle reported in Table 1 that would be the last to detach from the vascular wall? A. 0.2 pN B. 153.6 pN C. 68.7 Pa/m D. 121.9 Pa/m
Paragraph 3 hels to see that the largest minimum shear rate value is gc. the table shows that 2.20 s for vesicle N=33 degrees which has a lift forces of 153.6 pN.
According to the passage, all of the following are true about ankle injuries EXCEPT: A. they are more likely to occur in individuals of very low body weight who have been unable to build up ankle strength. B. a 13-mm translation of the talus relative to the tibia will cause injury. C. an anterior force of less than 98 N does not typically result in injury. D. torque of the talus increases the chance of ankle injury for a given applied force.
People who have low body weight would have less problems with ankle injuries because the passage states that the ankle injuries resort from people who are obese or overweight. Therefore, the answer is A.
According to the results presented by the experiment, the relationship between the lift force and the vesicle radius would best be described as: A. linear. B. inverse. C. exponential. D. sigmoidal.
Ran out of time for the last few. The 2nd and 3rd row of the columns show that the radius increases so this eliminates B. The slow increases at a high rate so therefore, C is the answer.
How many tetrahedral stereocenters are present in cholesterol (pictured below)? A. 2 B. 6 C. 8 D. 9
Sterocenters would be the same as chirality. therefore, C is the answer
A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30º. All of the following are true EXCEPT: I. the angle of reflection is 30º. II. the angle of refraction is 30º. III. total internal reflection will result, depending on the critical angle. A. I only B. I and III only C. II and III only D. I, II, and III
The angle of incident always equals the angle of refraction. The angle does not equal refraction because light enters a dense medium bends towards the normal. Since we knew I was true, the eliminates all answers except C.
Which of the following correctly identifies a chiral center in β-ODAP?
The answer is B, it has 4 different constituents attached to it (picture)
The data presented in Figure 2 most strongly supports which conclusion? A. β-ODAP effects on plasma methionine levels are strongest at the t = 2 to t = 3 hour marks. B. The sulfur-rich meal demonstrated no significant effect on elevating plasma methionine levels to baseline. C. The largest increase in methionine plasma levels after the sulfur-rich meal occurred between hours 11 and 12. D. The active compound in L. sativus seeds has a biological half-life of 1 hour.
The answer is B. The graph shows that there was no difference in met levels with the sulfur rich meal.
In laboratory experiments, the hydrolysis of ATP to ADP and inorganic phosphate has been shown to occur slowly. In the body, this process occurs rapidly. Which of the following may account for this observation? A. The action of enzymes present in the body, but absent in laboratory experiments, increases the proportion of ADP and inorganic phosphate present at the completion of the reaction. B. The hydrolysis of ATP is a non-spontaneous process that cannot proceed in the absence of enzyme present in the body, but not present in the laboratory experiment. Show Explanation C. Enzymes, present in the body but absent from the laboratory experiment, modify the standard free energy change of the hydrolysis of ATP, leading to a more favorable reaction profile. D. The catalytic activity of enzymes present in the body, but not in the laboratory experiment, reduce the activation energy associated with the hydrolysis of ATP.
The catalytic action of enzymes changes the activation energy and the reaction rate. I must've misread the question because it is clearly. D...
Given information provided in the passage, which of the following is the most likely standard voltage change for the ½ O2/H2O redox pair? A. +0.82 V B. +0.14 V C. -0.18 V D. -0.45 V
The final acceptor in the ETC is oxygen and in this problem, it is stated that E=0.22V and the only answer higher than that is A.
The enzyme listed in step 1 of the retinol synthesis listed is most likely classified as a(n): A. transferase. B. lyase. C. isomerase. D. oxidoreductase.
The final step in the equation is being oxidized. Therefore, the answer is D.
Which of the following reagents could be used to complete the final step of retinol synthesis shown in Figure 2? A. LiAlH4 B. O3 C.H2 with Pd D. KMnO4
The final step in the problem is a reduction of a retinal (aldehyde) to an alcohol. which would be a reduction reactive which gains hydrogen or oxygen. Therefore, A is the answer.
Which of the following cations will form most quickly from loss of a hydrogen atom?
The more energy required to remove a hydrogen atom from a compound equals the loss of a hydrogen atom. The more energy, the slower the rate. The answer is A because it has the lowest energy according to the table (pictures).
What is the correct order of the 5 para substituents on the carbocation intermediate, if arranged from most stabilizing to least stabilizing? A. F > H > CH3 > OCH3 > NO2 B. CH3 > NO2 > OCH3 > H > F C. NO2 > OCH3 > CH3 > H > F D. NO2 > CH3 > OCH3 > F > H
The more negative the RSE, the greater the stability. This was seen in table 1, which is put in order would make C the right answer.
Which of the following are products of the decay of I-131? A. Ionizing radiation, Xe-131, and an electron B. Ionizing radiation, Te-131, and an electron C. Non-ionizing radiation, Xe-131, and a positron D. Non-ionizing radiation, Te-131, and a positron
The passage says that I-131 undergoes B-minus decay which is adding an proton so that would be adding one to the atomic number. Gamma decay indicates ionizing radiation.
According to the experimental procedure, which of the following describes the physical properties of indium tin oxide? I. Opaque II. Electrically conducting III. Solid at standard temperature A. II only B. III only C. I and III only D. II and III only
The passage stated that the indium was clear which omits answer I and omits C. We know that it is electrically conducting because it is a metal and most experiments (unless stated otherwise) will have the metal solid at standard temperature.
When the researchers connected the solution-filled glass plates of the flow chamber to the AC generator, the ITO-coated plates mostly likely functioned as: A. a resistor. B. a capacitor. C. a galvanic cell. D. an electrolytic cell.
The passage states the the "plates were connectef to an AC generator and this indicates that there is a capacitor involved because voltage is involved and it was parallel-plate. Therefore, B is the correct answer.
Which of the following biological substances are likely derived from terpenes? I. Aldosterone II. Glucose III. Insulin IV. Estrogen A. I and II only B. I and III only C. II and III only D. I and IV only
The problem states that terpenes are steroids and the only two steroids on this list is aldosterone and estrogen. Therefore, D is the answer.
A researcher seeks to monitor the conversion of retinal to retinol using infrared spectroscopy. Which of the following will indicate the reaction is complete? A. The disappearance of peaks in the 3200-3500 cm-1 region B. The disappearance of a peak in the 1700-1750 cm-1 region C. The disappearance of a peak in the 1580-1640 cm-1 region D. The appearance of a peak in the 1700-1750 cm-1 region
The reaction of the -OH group would be complete by 3200-3500, so there would be no signal in 1700-1750 cm. Therefore, B is the correct answer.
During the experiment, scientists noted that several of the reaction beakers became hot to the touch. All of the following reactions could cause this result EXCEPT:
The reactions had to be exothermic in order for the heat to escape. Therefore, the answer should be endothermic. Hess's law is the Hf which would be products-reactants. Therefore, D is the correct answer.
The liquid remaining in the round-bottom flask at the end of the procedure was most likely: A. a mixture consisting of roughly even amounts of the two components. B. 2-methylundecanal. C. water condensed from the air in the lab. D. 2-methylundecanoic acid.
The substance that would be left would be one with a higher boiling point. Acids have a higher boiling point than aldehydes. Therefore, the answer is D.
According to the data in Table 1, what mass of captopril must be dissolved in 3 L of plasma at pH 7.4 to inhibit 50% of ACE enzyme activity in vivo? Assume an equal volume of distribution. A. 7.9 pg B. 7.9 ng C. 7.9 µg D. 7.9 g
The table shows that the PH range at 7.4 is 0.012um. The question tells us that plasma is 3 L so you would multiply 0.012 x10^-6 x 3 L plasma x 220 g= 7.92 um. If there's a 0 in the front, then multiply like you are multiplying the whole numbers. Therefore the answer is C.
Another possible method of separating 2-methylundecanal and 2-methylundecanoic acid could be based on: A. their differences in the rotation of plane-polarized light. B. a mass spectrometry analysis. C. an extraction based on their differing solubilities. D. the very different scent profiles of each molecule.
The two substances are different; one is an aldehyde (-al) and one is an alcohol (-ic, acid, -ate) so they would have to have different solubilities. Therefore, the answer would be C. The first answer would be if the two are enantiomers. The second one would be if you had to analyze the substances, not separate.
Tucson, Arizona is well known for its sunny, dry weather. A 33ºC day in Tucson can feel relatively cool and pleasant compared to a similar temperature in a humid city like Tampa. The phenomenon of "dry heat" feeling subjectively cooler is a result of: A. decreased evaporation from the skin helping to keep the body cool. B. increased evaporation from the skin helping to keep the body cool. C. decreased cardiac output in response to dehydration lowering cutaneous perfusion. D. increased cardiac output in response to dehydration lowering cutaneous perfusion.
The way we keep cool is to perspirate. The evaporation of liquid dissipates body heat, which would lower the body's temp and keep it from overheating. the dry climate has low humidity and allows water to
If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground? A. 325 N B. 650 N C. 750 N D. 1075 N
This one was a complete guess and I just so happened to get it right. In order to turn while running, your foot must push off the ground first while also supporting the weight of your body. So, the acceleration force must be considered. And this would be horizontal because the person is turning. We are ultimately looking for the hypotenuse. F(normal)= mg= 65 kg x 10 m/s= 650 N and F(turning)= 65 kg x 5 m/s2=325. Then, with rounding you would have to get the square root of 300^2+700^2. Then the answer would be between 700 and 800 leaving C as the correct choice.
According to passage information, is the energy released by the transport of an electron pair delivered by NADH through the electron transport chain sufficient to produce three ATP molecules from three ADP and three inorganic phosphate molecules if the efficiency of the conversion of the energy released via the transport of an electron delivered by NADH into useful work in the body is found to be approximately 50%? A. Yes, with the additional calories of energy released by the electron transfer and not used in the phosphorylation employed to drive the removal of excess hydrogens from the matrix. B. Yes, with the majority of calories of energy released by the electron transfer and not used in the phosphorylation lost mainly as heat. C. No, the energy released in the process is sufficient to drive the formation of only two ATP from two ADP and two inorganic phosphate molecules because Complex I is bypassed. Show Explanation D. No, the enthalpy change of the reaction is negative and the formation of ATP from ADP and inorganic phosphate requires the input of outside heat energy.
To answer, you need to find the amount of energy required to produce 3 molecules of ATP and the amount of energy released by the transport of NADH. Therefore, the answer is B.
Several families in areas 4 and 5 who read the results of the study believe that their children have no increased risk of developing thyroid cancer. Is this conclusion accurate? A. No, the mutations caused by I-131 exposure may take more than three years to manifest as cancer. B. No, I-131 has a half-life of only 8 days. C. Yes, Table 1 shows that there was no increase in cancer in areas 4 and 5 in years 2013 and 2014. D. Yes, the radiation given off by I-131 is not mutagenic.
Two part question/answer. Look at the table, and see that there is a fluctuant trend in these two area. So this would omit C&D. B can be omitted as well because the half life doesn't matter, the cancer can still take affect. Think of a person who stopped smoking cigarettes at a young age but develop cancer as an adult. Therefore, A is the answer.
Water is a rare substance in that the solid is less dense than the liquid at the freezing point, resulting in a solid form that floats on top of the liquid. Which of the following best explains this phenomenon? A. The bent structure of the water molecule results in a molecular dipole that maximizes the close molecular packing structure in the solid state. B. The bent structure of the water molecule and ratio of covalently-bonded hydrogens to lone pairs of electrons on the oxygen atom maximizes the hydrogen bonding that occurs in the solid phase, producing a hexagonal structure with large empty spaces. C. The degree of ionization in the solid state is less than in the liquid phase. D. The London dispersion forces of water significantly decrease in the solid phase as compared with the liquid phase.
We know that water is a bent structure so we can exclude C and D. Also know that water has hydrogen bonding. So the only option would be B.
During the anterior drawer test, the joint translation occurs in 0.2 s. What is the average velocity of the talus during translation? A. 2.85 x 10-3 m/s B. 2.85 x 10-2 m/s C. 3.0 x 10-2 m/s D. 3.0 m/s
velocity= d/t (distance/time). Distance is 5.7 mm or 6 mm. then divide this by 0.2 s= 30 mm/s= 3 x10^-2. With mental math and rounding, the answer would be B. With the decimal go up to a whole number for example 6/0.2 = 30 and 10/0.5 would be 20
Which of the following protonation states of the captopril thiol and carboxyl groups is required to maximize captopril's inhibition of ACE? A. Deprotonated thiol and protonated carboxyl B. Protonated thiol and deprotonated carboxyl C. Deprotonated thiol and deprotonated carboxyl D. Protonated thiol and protonated carboxyl
With this one I guessed because I didnt really understand the question. This would require looking back into the table to see that captopril thrives in environments that are 3.5-9.5. Therefore, there would be a deprotonated thiol because pKa would be lower than 9.8 and a protonated carboxyl because the pKa is greater than 3.7. Therefore, the answer is B.
