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MCAT

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MCAT- Prep

© Prep101 http://www.prep101.com/mcat/

Passage I bone mass, while a lack of activity


activity allows for
further bone loss.
Human bone is composed of two portions:
1) a hard mineral portion made up primarily of 1. In order to maintain normal bone tissue, the
calcium; and 2) a softer, organic, collagen-like body would have
have to increase the level of
matrix. 99% of the body’s calcium is contained in calcitonin in response to what condition?
bone tissue. Bone is a dynamic
dynamic substance,
continuously being resorbed and reformed A) The level of calcium in the plasma is high
throughout the lifespan of a person. This renewal B) The level of parathyroid hormone is too low
process is directly related to the maintenance of an C) There is a dietary deficiency of calcium
adequate level of calcium in the blood stream. D) There is a dietary deficiency of vitamin D
There are several important molecules that affect
this relationship, such as vitamins C and D, 2. Which of the following persons would be
calcitonin and parathyroid hormone. most likely to have osteomalacia?

Vitamin C is required for the formation and A) A child with a dietary deficiency in fat-
maintenance of connective tissue, including soluble vitamins living in a tropical climate

r
synthesis of bone matrix. A deficiency in vitamin B) A child with a dietary deficiency in fat-

se
C results in scurvy. Although the effects of soluble vitamins living in a northern climate
calcitonin are minute, it acts to decrease bone C) A child with a dietary deficiency in water-
resorption, helping to precisely calibrate bone soluble vitamins living in a tropical climate
quantity and arrangement. Parathyroid hormone ta D) A child with a dietary deficiency in water-
stimulates formation and activity of osteoclasts, soluble vitamins living in a northern climate
which break down bone cells, and impair new
bone formation. Vitamin D actsacts as a hormone in 3. When taking a calcium supplement,
its activated form, stimulating absorption of nutritionists recommend that it include
on
calcium by the small intestine. Vitamin D also vitamin D because
acts on bone tissue to enhance the effect of
parathyroid hormone. Although
Although vitamin D can be A) the activated form of vitamin D enhances the
obtain from diet, it is one of the few hormones uptake of calcium by bone tissue.
humans can successfully synthesize themselves B) vitamin D is needed to prevent osteomalacia.
M

when skin is exposed to ultraviolet light. C) the activated form of vitamin D enhances the
action of calcitonin.
There are two common diseases that affect bone D) the activated form of vitamin D stimulates the
r.

metabolism, osteomalacia and osteoporosis. absorption of calcium into the blood.


Osteomalacia generally affects children more
often than adults and is known as rickets in 4. Which of the following circumstances would
D

children. A deficiency in vitamin D causes the likely result in a case of osteomalacia?


inadequate mineralization of new bone matrix and
results in a lower ratio of mineral to organic A) Metabolic deficiency of parathyroid hormone
matter in bone tissue than would be observed B) Metabolic deficiency of parathyroid hormone
normally. Many suffers have bowed legs, as this and impairment of conversion of vitamin D to
disease leads to distortion of the bones, especially its active form
the long bones. Osteoporosis, on the other hand, is C) Metabolic deficiency of parathyroid hormone
mainly a disease of the elderly, striking and an inability of the active form of vitamin
postmenopausal women most most often. More bone is D to acts on its target tissue
resorbed than formed, leading to a reduction in D) Impairment of conversion of vitamin D to its
bone mass. The remaining bonebone has a normal ratio active form and an inability of the active form
of mineral to organic matter content. Scientists of vitamin D to acts on its target tissue
have yet to understand why the body loses bone
mass. However, studies show that an increase in
weight-bearing activity promotes increases in

Page 1 of 17 (STEM-2021)
MCAT- Prep
5. Hypocalcaemia (low levels of calcium in the Passage II
blood) will trigger an increase of:
Norman, born and raised in South Carolina, is
A) osteoclast activity. training to become a U.S. Marine. Over the course
B) osteoclast activity and vitamin C. of his one-year training course, Norman makes
C) osteoclast activity and parathyroid hormone. several trips to different locations to learn new
D) parathyroid activity and vitamin C. skills.

6. A tumour necessitated the removal of the Shortly after joining the training program,
parathyroid gland. What effect would the Norman’s unit went to Florida to learn to skin
post-operative patient experience? dive. Norman, being a marine-in-training, was in
excellent physical condition. However, the first
A) A drastic change in the ratio of mineral to time he went diving, Norman noticed that he
matrix tissues in bones experienced elevated pulse and respiratory rates.
B) An increase in calcitonin production to Norman performed several more dives over the
compensate for calcium deficiency in the course of the week, and by the third dive, his heart
plasma and breathing rates were no longer elevated. By
C) Calcification of some organs due to the end of the week, his skin had become darker.
accumulation of calcium in the plasma

r
D) Severe neural and muscular problems due to Several months later, Norman’s unit went to the
deficiency of calcium in the plasma Rocky mountains to learn to ski. Once again,

se
Norman noticed that the first time he went skiing,
7. Early migration of humans into more his heart and ventilation rates were faster than
northerly latitudes, such as Europe, altered the normal. By the end of the week, although there
environmental factors humans were exposed ta was a decrease in heart and breathing rates, the
to. To prevent diseases such as osteomalacia, rates were still slightly elevated. Norman also
anthropologists hypothesize that: noticed that his appetite had increased over the
course of the week and that he had significantly
A) the amount of melanin in the skin decreased increased his food intake but this had no effect on
on
while consumption of dairy products his weight. Norman didn’t notice the same
increased. magnitude of increase in his food intake during
B) the amount of melanin in the skin decreased the diving trip.
as did the consumption of dairy products.
M

C) the amount of melanin in the skin increased Norman calculated the actual amount of work that
while consumption of dairy products he performed during the skiing and diving trips
increased. and found that there was not enough difference in
r.

D) the amount of melanin in the skin increased as the work performed to account for the observed
did the consumption of dairy products. difference in appetite. The physical work of
diving and skiing were approximately equal but
D

8. Which of the following is not an expected the amount of calories ingested during the skiing
symptom of scurvy? trip were greater than the diving trip.

A) Bleeding under the skin


B) Fluctuating blood pressure and heart rate
C) Tingling sensation in the extremities
D) Brittle bones

9. Which is not a function of the skeleton?

A) Mineral storage
B) Protection of vital organs
C) Production of blood cells
D) Detoxification of poisons.

Page 2 of 17 (STEM-2021)
MCAT- Prep
10. Control of heart rate, muscle coordination, 14. During the initial skin diving session, when
and appetite is maintained by the: his heart and breathing rates were elevated,
Norman noticed that he produced more urine
A) brain stem, cerebellum, and hypothalamus than usual. This was probably due to:
respectively
B) brain stem, hypothalamus, and cerebrum A) inability to cool the skin through evaporative
respectively water loss.
C) hypothalamus, cerebrum, and the brainstem B) increased blood pressure caused by
respectively excitement or anxiety.
D) cerebellum, hypothalamus, and brain stem C) reduced blood pressure caused by excitement
respectively or anxiety.
D) absorption of water from the ocean.
11. Norman noted that even though his skin blood
vessels were usually constricted to conserve 15. The effect of the parasympathetic autonomic
body heat in the cold environment of the nervous stimulation on the heart is:
mountains they would occasionally dilate for
short periods of time. What would be the most A) to release adrenaline and excite the heart.
probably physiological purpose for this B) to release adrenaline and slow the heart.
periodic vasodilation? C) to release acetylcholine and excite the heart.

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D) to release acetylcholine and slow the heart.
A) Maintain normal muscle tone

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B) Maintain normal skin tone 16. When Norman was skin diving under the
C) Maintain sufficient oxygenation of cells water, he could clearly hear the “lub-dub” of
D) Reduce excessive blood pressure his heart beating? The “lub” sound the heart
ta makes is caused by the:
12. The initial increase in heart and breathing
rates during the skin diving training was A) closing of the mitral and tricuspid valves.
probably due to: B) closing of the pulmonary and aortic valves.
C) sound of blood rushing into the atria.
on
A) elevated core body temperature caused by D) sound of blood rushing into the ventricles.
swimming in warm water.
B) activation of the parasympathetic autonomic
nervous system by the new experience. 17. The darkening of the skin during tanning:
M

C) activation of the sympathetic autonomic


nervous system by the new experience. A) is naturally healthy and is a sign of vigor in
D) hypoxia caused by the inability to supply humans
r.

sufficient oxygen for the strenuous exercise of B) is a direct response to exposure to visible light
swimming at sea level. C) protects the dermis from damage by
increasing the number of pigment-producing
D

13. The prolonged increase in heart rate and cells


breathing rates during the snow skiing trip D) protests the dermis from damage by
was probably due to: increasing the production of pigment in
melanocytes.
A) depressed core body temperature
(hypothermia) caused by exposure to cold
temperatures at high altitude.
B) hypoxia caused by insufficient blood
hemoglobin concentration to supply oxygen
for exercise at the low oxygen pressure found
at high altitudes.
C) activation of the parasympathetic autonomic
nervous system by the new experience.
D) activation of the sympathetic autonomic
nervous system by the new experience.

Page 3 of 17
MCAT- Prep
Passage III resurgence since the discovery that the bacterium
Helicobacter pylori may play a causal role in
Inflammatory bowel disease (IBD) is a group of gastric ulcers.
inflammatory conditions of the large, and
sometimes small, intestine. Crohn’s disease (CD)
and ulcerative colitis (UC) are the two most 18. An ulcer that penetrated the wall of the
common forms of IBS. The main difference intestine would allow the contents of the
between CD and UC is the location and nature of gastrointestinal tract to enter:
the inflammatory changes in the gut. CD can
affect any part of the gastrointestinal tract, from A) the pleural cavity.
mouth to anus. UC, in contrast, is restricted to the B) the peritoneal cavity.
colon. CD usually involves all layers of the C) the lumen of the intestine.
digestive tract while UC affects only the mucosa D) the perineum.
(inner lining of the tract). The first symptoms
usually experienced by a suffer of CD are 19. What process would be most disrupted by an
abdominal pain and diarrhea following meals, inflammation of the colon?
while a progressive loosening of a bloody stool is
the first symptom of UC. A) Absorption of nutrients
B) Absorption of water

r
Management of IBD is achieved by drug therapy C) Digestion
to suppress the inflammation which leads to D) Secretion of digestive enzymes

se
diarrhea, but there is no known cure. The cause of
IBS is controversial, with genetic, pathogenic, and 20. Normally the immune system avoids attacking
immunogenic theories all having been advanced. the tissues of its own body because:

IBD tends to run in families, with 20% of patients


having a relative with the disorder. The incidence
ta A) the body does not make any antigens that the
immune system could recognize.
of CD in North America and Europe is 1 in B) it changes its antibodies to be specific only to
600,000 but are lower in Africa and Asia. foreign antigens.
on
Incidence rates are higher in northern latitudes of C) a special intracellular process recognizes only
northern countries and in particular ethnic groups. foreign antigens.
If IBS is genetically linked, it is not inherited in a D) it suppresses cells specific to the body’s own
simple Mendelian manner. antigens.
M

Some research suggests that IBD is an 21. The fact that there appears to be a genetic
autoimmune disease. An antigen in the body, component to IBD which does not
r.

perhaps in the digestive tract, is recognized as demonstrate Mendelian inheritance patterns,


foreign by the immune system. This antigen may suggests that:
then stimulate the body’s defenses to produce an
D

inflammatory response that continues without A) the disease is polygenic.


control. Recently the “helminthic theory” was put B) the gene for the disease is recessive.
forward, suggesting that a lack of traditional C) the gene for the disease has incomplete
targets, parasites such as roundworms and dominance.
whipworms, stimulates or perhaps increase this D) the gene for the disease has limited
autoimmune response. Rates of IBD have expressivity.
increased in countries where in parasitic infection
rates have decreased.

There is also a lingering suspicion that the


inflammation is triggered by some bacterium or
other organism that takes up residence in the
gastrointestinal system. Mycobacterium
paratuberculosis, a bacteria found in cows, sheep
and goats, causes Johne’s disease which has many
similarities to CD. This theory has also had a

Page 4 of 17
(STEM-2018)
MCAT- Prep
22. If the genetic and autoimmune theories of Passage IV
IBD are true, then the gastrointestinal antigen
being targeted by the immune system is A set of experiments was carried out to isolate
probably on: unusual mutants in the bacterium Staphylococcus.
It is known that normal staph cells grow well on a
A) the chromosomes carrying the genes for the minimal nutrient medium containing agar and a
disease. solution of glucose, salts, and ammonia. Staph can
B) the surface of the proteins encoded by the also grow on a medium containing more complex
genes for the disease. sugars like lactose, because they normally have
C) part of the DNA segments constituting the the enzymes needed to convert complex sugars to
genes for the disease. glucose. Overnight, a single bacterial cell can
D) Stretches of the mRNA’s coded for by the grow into a culture of millions of cell that appears
genes for the disease. as a raised colony on the medium.

23. Which of the following would make you Experiment 1


question the involvement of a bacterial
infection in UC? A sample of normal staph was treated with X-rays
in order to produce genetic mutations in the
A) The presence of worms in the gut of the bacterial cells. The bacteria were then spread

r
patient. across a dish containing a minimal nutrient media.
B) The similarity of the ulcers to that of gastric The medium also contained the antibiotic

se
ulcers. ampicillin, which kills normal staph bacteria. The
C) The presence of similar conditions in other culture was incubated overnight and examined.
mammals. No bacterial cultures were found growing in the
D) The presence of symptoms outside the ta Petri dish.
gastrointestinal tract.
Experiment 2
24. The helminthic theory holds that:
The experiment was repeated exactly as before,
on
A) roundworm infection causes IBD using the same original sample. This time when
B) a genetic protection against IBD is triggered the culture was examined after incubation, two
by the presence of roundworms colonies of bacteria (Colony A and Colony B)
C) Mycobacterium paratuberculous causes IBD were found growing on the medium.
M

D) the absence of roundworms increases the


autoimmune response of the body. Experiment 3
r.

25. Crohn’s disease is different from ulcerative Samples from both colonies grown in Experiment
colitis because 2 were spread across two different ampicillin-
containing media. Medium X was the typical
D

A) CD affects the entire GI tract and is probably minimal medium while Medium Y contained the
caused by a bacteria. same ingredients, with the exception that instead
B) CD affects only the mucus membrane of the of glucose, lactose was the only nutrient sugar
colon and is probably caused by a bacteria. present. After incubation, the four dishes were
C) CD affects the entire GI tract and is not examined and the results are shown in Table 1.
treatable by surgery.
D) CD affects only the mucus membrane of the Table 1. Results of Experiment 3
colon and is not treatable by surgery.
Medium X Medium Y
Colony A growth growth
Colony B growth no growth

Page 5 of 17
(STEM-2019)
MCAT- Prep
26. What conclusions can be drawn from 30. If ampicillin-resistance arose by mutation in
Experiment 1? Experiment 2, when did the mutation that
prevented Colony B from growing in the
A) All bacterial cells became ampicillin-resistant. lactose medium arise?
B) No ampicillin-resistant mutations were
produced. A) in Experiment 1
C) Mutations were produced, but none that were B) in Experiment 2
ampicillin-resistant. C) in Experiment 1 or Experiment 2
D) No mutations were produced. D) in Experiment 3

27. What conclusions can be drawn from 31. If a culture taken from the original x-rayed
Experiment 2? sampled in Experiment 1 was spread across a
fresh dish of minimal medium without
A) All bacterial cells except two became antibiotics, what would you expect to observe
ampicillin-resistant. after incubation?
B) At least two bacterial cells were mutated and
became ampicillin-resistant colonies. A) The cells that produced Colony A in
C) Colony A and Colony B can grow without Experiment 2 would die on the medium.
glucose in the medium. B) Only cells that produced Colony A in

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D) A variety of mutants were produced with the Experiment 2 would be able to grow.
second dose of X-rays. C) One or two colonies would be growing on the

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medium.
28. What do the results of Experiment 3 suggest D) Colonies would be growing everywhere on
about how Colony A and Colony B are the medium.
similar to each other? ta 32. Which of the following would not be a
A) Both can grow on glucose, lactose, and mechanism by which cells in Colony A could
ampicillin. convey their ampicillin-resistance to other
B) Both can grow on glucose and lactose. cells?
on
C) Both can grow on glucose and ampicillin.
D) Both can grow on lactose and ampicillin. A) Transduction
B) Conjugation
29. What might be a reasonable hypothesis for C) Transformation
M

explaining why Colony B is unable to grow D) Transference


on Medium Y?
r.

A) The gene for the enzyme that converts lactose 33. Which statement about antibiotics and
to glucose had mutated. antibiotic drug use is false?
B) Lactose is not nutritious when mixed with
D

ampicillin. A) Most antibiotics are non-specific, also killing


C) A gene in the cells that made up Colony A beneficial bacteria that normally occur in the
had mutated, enabling them to out-compete body.
the cells of Colony B in the lactose medium. B) Antibiotic use can accelerate the evolution of
D) The mutagenic X-rays affected the lactose in antibiotic resistance in diseases such as
the medium. tuberculosis.
C) Antibiotics accelerate the rate of mutation in
the targeted bacteria, causing resistance to
appear in the bacteria.
D) Antibiotic-resistance genes can move between
different species of bacteria through a process
called “horizontal gene transfer.”

Page 6 of 17
(STEM-2019)
MCAT- Prep
Passage V 34. Enzymes A and B in the figure are both
protein-digesting enzymes found in humans.
Enzymes act as biological catalysts that help Where would they most likely be found?
speed up reactions by lowering the activation
energy needed to bring reactants to their A) A in the small intestine; B in the stomach
“transition state.” In this unstable condition, bonds B) A in the mouth; B in the small intestine
can break and the reaction proceeds. The reaction C) A in the stomach; B in the small intestine
itself is often described in the following way: D) A in the small intestine; B in the mouth

E + S  ES  P + E 35. Which statement is true concerning enzymes


A and B?
E = enzyme ES = enzyme-substrate
complex A) They could not possible be at work in the
S = substrate(s) P = product(s) same part of the body.
B) At a pH of 4.5, enzyme A works slower than
Enzymes can carry out their catalytic functions enzyme B.
only with particular substrates, and only under C) They have different temperature ranges at
particular environmental conditions (such as pH which they work best.
and temperature). This characteristic of enzymes D) At their appropriate pH ranges, both enzymes

r
is referred to as enzyme specificity. The enzyme work equally fast.
specificity in four different enzymes is shown in

se
the Figure 1. 36. What conclusion may be drawn concerning
enzymes X and Y?

ta A) Neither enzyme is likely to be a human


enzyme.
B) Enzyme X is more likely to be a human
enzyme.
C) Enzyme Y is more likely to be a human
on
enzyme.
D) Both enzymes are likely to be human
enzymes.
M

37. At which temperatures might enzymes X and


Y both work?
r.

A) Above 40ºC
B) Below 50ºC
C) Above 50ºC and below 40ºC
D

D) Between 40ºC and 50ºC

38. An enzyme-substrate complex can form when


Figure 1. Activity of four enzymes at varying
the substrate(s) binds to the active site of the
temperature and pH.
enzyme. Which environmental condition
might alter the conformation of an enzyme in
the figure to the extent that its substrate is
unable to bind?

A) Enzyme X at 40ºC
B) Enzyme A at pH 4
C) Enzyme Y at pH 2
D) Enzyme B at 37ºC

Page 7 of 17
(STEM-2019)
MCAT- Prep
39. At 35ºC, the rate of the reaction catalyzed by Passage VI
enzyme X begins and levels off. Which
hypothesis best explains this observation? Muscular dystrophy (MD) is one of the most
common genetic diseases found in North America
A) The enzyme has become saturated with with one in every 4000 boys (and rarely in girls)
substrate. suffering from the degenerative disease. Although
B) The temperature is too far below optimum. normally inherited, it is estimated that 30% of
C) Both A and B cases are due to a spontaneous genetic mutation.
D) Neither A nor B MD results in the progressive degeneration of
skeletal and cardiac muscle fibers, weakening the
40. In which of the following environmental muscle and leading ultimately to death from either
conditions would digestive enzyme B be cardiac or respiratory failure. Duchenne muscular
unable to bring its substrate(s) to the transition dystrophy (DMD) is the most common form of
state? MD, affecting children at birth. However, some
people are struck with similar symptoms later in
A) At any temperature below optimum. life, with a related, but milder, form of MD known
B) At any pH lower than 5.5. as Becker’s muscular dystrophy (BMD).
C) At any temperature higher than 37ºC
D) At any pH where the rate of reaction is not DMD is caused by a mutation in the gene which

r
maximum. codes for the protein dystrophin, rendering it
nonfunctional. Also the result of a mutation, BMD

se
41. If the change in heat of a chemical reaction is is usually characterized by a partially functional
negative and the change in entropy is positive, dytrophin product, accounting for its later onset.
what can you conclude about the reaction? The protein dystrophin is found on the inner
ta surface of the plasma membrane in normal muscle
A) It requires energy. tissue. The gene that codes for dystrophin is the
B) It is endergonic. longest known gene in the human body at 2.4
C) It is spontaneous. megabases in length, accounting for 0.1% of the
D) It will not reach equilibrium. genome.
on

42. Which statement most completely describes Therapeutic alteration of the genome as a
the nature and function of enzymes? treatment for genetic disorders (gene therapy)
poses formidable challenges. Among them is the
M

A) Enzymes are molecules that are involved in problem of targeting altered genes to the correct
digestion. cells. One method of introducing genes into cells
B) Enzymes are proteins produced by living is to coat the DNA with an envelope of cationic,
r.

organisms that facilitate chemical reactions. amphipathic lipids. The packets formed in this
C) Enzymes are proteins in bodily secretions that way are called liposome-DNA complexes. The
protect the body from infection. lipids present their charged surface to the DNA
D

D) Enzymes are inorganic chemicals that act as molecules giving them an outer hydrophobic
catalysts to increase the rate of a reaction, surface coating which enables the DNA to cross
without being used up themselves. the plasma membrane.

43. Which of the following could result in


negative feedback of enzyme activity?

A) A substrate molecule binding to the active site


in the enzyme.
B) A lack of substrate available for catalysis.
C) A product molecule binding to an allosteric
site on the enzyme.
D) The product of a reaction destroying the
enzyme.

Page 8 of 17
(STEM-2019)
MCAT- Prep
44. The liposome-DNA complex crosses a 49. The functions of dystrophin likely include all
membrane composed of: but which of the following?

A) lipids and proteins. A) Protection of elements within the membrane


B) lipids and DNA. during contraction of cells
C) lipids, DNA, and proteins. B) Maintenance of the structural integrity of the
D) lipids only. plasma membrane
C) Keeping ion channels within the cells open
45. Which of the following best explains the high D) Recognition of protein hormones important to
incidence of muscular dystrophy in males? the functioning of the cells

A) Fathers, who have the disease, pass it on their 50. The high rate of spontaneous mutation in the
sons, but not to their daughters. dystrophin gene is probably due to:
B) Mothers, who are carriers of the disease, pass
it on to their sons, but not to their daughters. A) the folding of the gene into the chromosome
C) The disease is sex-linked and the gene which allows for transposition of nearby bases.
codes for dystrophin occurs on the Y B) the long length of the gene is statistically
chromosome. more likely to undergo point mutations.
D) The disease is sex-linked and the gene which C) the presence of another gene affecting the

r
codes for dystrophin occurs on the X dystrophin gene.
chromosome. D) X-inactivation results in only one copy being

se
available to be inherited by offspring.
46. Lipids used to complex with DNA must be
cationic because DNA is: 51. Which of the following signs would be
ta expected when diagnosing DMD?
A) negatively charged and hydrophobic.
B) negatively charged and hydrophilic. A) Abnormally enlarged muscle fibers
C) positively charged and hydrophobic. B) Decreased levels of creatinine kinase in the
D) positively charged and hydrophilic. muscle cells
on
C) A normal electrocardiogram
47. Targeting therapeutic genes to specific cells is D) Increased levels of dystrophin in the muscle
important because: tissue
M

A) all cells contain and express the same genes 52. Analysis of the dystrophin gene obtained from
B) all cells contain, but do not express, the same numerous patients identified large deletions,
genes often in different locations. One of the most
r.

C) inherited defective genes are only found in common deletions occurred in the center of
some cells of the body the gene. The sequence of this segment was
D) different cells contain different genes determined to be:
D

48. Would an exogenous gene introduced into 5’--GCCATAGAGCGA--3'


somatic cells be passed from parent to
offspring? Which of the following sequences of DNA is
complementary to this sequence?
A) No, because changes to somatic cells are not
passed down A) 5’--TCGCTCTATGGC--3'
B) No, because foreign DNA is not passed to B) 5’--CGGUUUCUCGCU--3'
offspring C) 5’--CGGTATCTCGCT --3'
C) Yes, because the DNA will be localized into D) 5’--CGGUAUCUCGCU--3'
the cell’s nuclei
D) Yes, because all DNA is passed from parent
to offspring during meiosis

Page 9 of 17
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MCAT- Prep
Biology Action Potentials Passage III

Passage III (Questions 14-20) If a stimulus is strong enough so that the threshold
potential of a cell is reached (about 15 mV from the Em),
Many animal cells have an intracellular fluid (ICF) an action potential (Figure 2) will be generated that will
that is electrically negative relative to the extracellular propagate along the nerve cell until it reaches the
fluid (ECF), creating a voltage difference across the cell's presynaptic ending. A neurotransmitter is released and
membrane called the resting membrane potential (Em). diffuses across the synaptic cleft, where it binds to the
The Em in nerve cells and muscle cells is about -80 mV, postsynaptic membrane. A postsynaptic potential of equal
while in epithelial and red blood cells it is about -30 mV. magnitude is generated in the postsynaptic cell and the
signal continues. The response of the action potential is
In nerve and muscle cells the concentration gradients all-or-none. The phases are shown in Table 2.
and the permeability of the membrane to both Na® and
K® determine the Em. Theconcentrations of these ions in
the ICF and ECF are shown in Table 1. They are
maintained by the Na®/K®-ATPase pump, a pump that

r
transports 3 Na® ions to the ECF for every 2 K® ions
transported to the ICF. At rest, the relative permeability
of the membrane to K® is roughly 10times greater than it

se
is to Na®.

Table 1
Ion [Intracellular] [Extracellular] tEi( ta Time (msec)
Na+ 15 mM 140 mM +58 mV

K+ 135 mM 4mM -92 mV Figure 2


'Ejon = Equilibrium potential for the ion
Table 2
on
1. Resting State: Thecell at itsEm.
If the Em transiently changes from its resting value, 2. Threshold: Potential that generates an AP.
an electrical signal is generated in the nerve cell. There 3. Depolarization: Due to influx of Na+.
are two types of electrical signals: graded potentials and
4. Overshoot: AP with a positive membrane potential.
action potentials (AP). Graded potentials are signals that
operate over short distances, while action potentials are 5. Repolarization: Due to efflux of K+.
M

signals that operate over long distances. 6. Hyperpolarization: Membrane potential is more
negative thanthe E^
Along the plasma membrane of a nerve cell are 7. Absolute Refractory Period: AP cannot be generated.
voltage-sensitive Na® and K® channels that contain
r.

8. Relative Refractory Period: AP can be generated if the


gates. TheNa® channel has anmgate that opens quickly stimulus is strong enough.
and an h gate that closes slowly upon depolarization. The
K® channel has an n gate that opens slowly upon
D

depolarization. The Na® gate can be blocked by


tetrodotoxin (TTX), and the K® gate can be blocked by
tetraethylammonium (TEA). In the resting state the Na®
and K® channels have their gates arranged as shown in
Figure 1: 14. The threshold potential for the generation of an
action potential can be found at that value of the Em
where the:
m ** ECF n
A. influx of K® balances the efflux of Na®.
B. efflux of K® is 10times greater than the influx
I j membrane | ofNa®.
1/\l icf 7 C. efflux of K® balances the influx of Na®.
K+ D. influx of Na® is 10 times greater than the
efflux of K®.
Figure 1

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Specializing in MCAT Preparation
MCAT- Prep
Biology Action Potentials Passage m

15. The threshold potential for the action potential 20. Gamma-aminobutyric acid (GABA) is a major
shown in Figure 1 is: inhibitory neurotransmitter in the central nervous
system. It is synthesized from glutamate, a major
A. -50 mV. excitatory neurotransmitter in the brain.
B. -65 mV.
C. -80 mV. COOH
D. -92 mV. © i © i
HjN-C-H HjN-C-H
Enzyme
CH, CH2
I
16. During depolarization, the quick opening of the CH, CH,
I I
voltage dependent m gates results in all of the COOH COOH
following EXCEPT:
Glutamic acid GABA

A. an influx of Na® ions.


B. an increase in the opening of fast m gates in The enzyme that catalyzes this reaction is best

r
described as:
neighboring Na® channels.
C. an increase in the opening of fast n gates in
A. GABA decarboxylase.

se
neighboring K® channels. B. glutamic acid decarboxylase.
D. an absolute refractory period. C. GABA carboxylase.
D. glutamic acid carboxylase.

17. Which of the following voltage gated positions best


represents the repolarization phase?

A.
B.
m gate open; h gate closed; n gate open.
m gate closed; h gate closed; n gate open.
ta
C. m gate open; h gate open; n gate open.
on
D. m gate closed; h gate open; n gate closed.

18. During the last half of the relative refractory period


shown in Figure 1, the membrane potential is
beginning to:
M

A. repolarize.
B. hyperpolarize.
C. depolarize.
D. hypopolarize.
r.

19. If TTX and TEA are added to a nerve cell


preparation and a depolarizing stimulus is applied to
D

the presynaptic ending, a postsynaptic potential will


be produced. This shows that:

A. Na® and K® ions are needed for the release of


the neurotransmitter to occur.
B. Na® and K® ions are not responsible for the
release of the neurotransmitter.
C. TTX and TEA can amplify the postsynaptic
potential by allowing neurotransmitter release.
D. an increase in ICF K® ions and an increase in
ECF Na® ions inhibit the release of the
neurotransmitter, unless a depolarizing
stimulus is applied.

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MCAT- Prep
Biology The Lens, the Iris, & Associated Muscles Passage V

Passage V (Questions 27-33) 27. For a person watching a meteor shower at midnight,
what are the contraction states of the circular muscle
Figure 1 shows the lens, the iris, and the muscles of of the iris, the radial muscle of the iris, and the
the eye. ciliary muscle?

A. The circular muscle of the iris is contracted,


the radial muscle of the iris is relaxed, and the
Pupil ciliary muscle is relaxed.
Cornea B. The circular muscle of the iris is relaxed, the
radial muscle of the iris is contracted, and the
ciliary muscle is contracted.
Suspensory C. The circular muscle of the iris is relaxed, the
Ligaments radial muscle of the iris is relaxed, and the
Optic ciliary muscle is relaxed.
Nerve
D. The circular muscle of the iris is relaxed, the
radial muscle of the iris is contracted, and the

r
ciliary muscle is relaxed.
Figure 1

se
The opening of the pupil of the eye is controlled by
two sets of muscles. The circularly arranged smooth
muscle is under parasympathetic control, and the radially
arranged smooth muscle is under sympathetic control.
Figure 2 shows the relation between these muscles and the
pupil of the eye. The size of the pupil reacts to the amount
of light present.
ta 28. As people age, the lens becomes less llexible and
therefore less able to change shape during
accommodation. What happens to a person's vision
with these age changes?

Circular Smooth A. The lens is less able to focus the light rays
on
Radial Smooth
Muscle Fibers Muscle Fibers from near objects.
B. The lens is more able to focus the light rays
from near objects.
C. The lens is more able to focus the light rays
from far objects.
D. The lens is less able to focus the light rays
M

from far objects.


r.

Dim Light Bright Light 29. Special solutions are used to dilate the pupil during a
D

retinal exam, so that the retina can be viewed. The


Figure 2 solution causes an inability to focus on near objects,
reduces clear distance vision, and enlarges the pupil.
The lens of the eye is suspended within and What is the stimulation or inhibition required to
surrounded by a ring of tissue called the ciliary muscle. produce these changes in the eye?
Suspensory ligaments connect the lens to the ciliary
muscle. When the ciliary muscle is relaxed, the A. Sympathetic nerves arc inhibited, and
suspensory ligaments pull the lens taut and Hat. Since we parasympathetic nerves are stimulated.
spend most of our time in far-away vision (20 feet or B. Sympathetic nerves are stimulated, and
more), this is ideal. For closer work, the lens must parasympathetic nerves are inhibited.
accommodate and become thicker for focusing. In the C. Both are stimulated.
process, the ciliary muscle tightens, the suspensory D. Both are inhibited.
ligaments become slack, and the lens thickens and
becomes more convex.

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Biology The Lens, the Iris, & Associated Nusdes Passage V

30. When light rays travel through the pupil and the
lens, upon which part of the eye does the image
focus?

A. Retina
B. Cornea
C. Optic nerve
D. Vitreous body

31. One eye exercise involves focusing on an object


while it moves from arm's length to the closest point
upon which you can focus. This causes the lens to
undergo accommodation, that is, the lens must
change shape to keep the object in focus as it moves

r
closer. As the object moves nearer, what changes
occur in the eye?

se
I. The pupil gradually contracts.
II. The ciliary muscle gradually contracts.
III. The suspensory ligaments gradually contract.

A.
B.
C.
D.
I only
II only
IH only
II and III only
ta
on
32. The iris is pigmented epithelial tissue. The color of
the eye is determined by the amount of pigment.
Blue eyes have the least pigment, brown eyes have
more, and black eyes have the greatest amount of
pigment. What color are the eyes of a person who
M

has no pigment in her iris?

A. Black
B. White
r.

C. Pink
D. Green
D

33. Astigmatism is the condition of having a non-


uniformly shaped lens or cornea. This means that
parallel light rays do not focus, so that a sharp image
is not formed. How could this eye condition be
corrected?

A. By providing a convex corrective lens.


B. By providing a concave corrective lens.
C. By providing a non-uniform corrective lens.
D. By providing a uniform corrective lens.

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Biology Gregor Mendel and Inheritance Passage II

Passage II (Questions 6-11) 6. Mendel first allowed plants of a given variety to


produce progeny by self-fertilization over many
The first quantitative studies of inheritance involved generations in order to:
the garden pea and were carried out by Gregor Mendel.
His choice of the pea was fortuitous, because a large A. carry out an experimental cross,
number of true-breeding varieties (uniform from one B. assure himself the plants were of the true-
generation to the next) was available. Mendel selected breeding variety.
seven easily distinguishable traits of these plants (e.g., C. observe segregation of alternative traits among
purple versus white flowers, and round versus wrinkled progeny,
seeds). D. carry out a reciprocal test-cross.

Pea plants contain both male and female sex organs.


This was also advantageous to Mendel, because he could
either let self-fertilization take place within an individual
flower or perform an experimental cross. A cross involves

r
removal of a flower's male parts before fertilization can
occur and introducing pollen from a strain with alternative
Mendel discovered that out of the F2 generation for

se
characteristics. 7.
any trait, 1/2 of the individuals were not true
Mendel first allowed plants of a given variety to breeders. This class of plants, if allowed to self-
produce progeny by self-fertilization for many pollinate, should produce Fj individuals that exhibit
generations. He then conducted crosses. Mendel took a dominant and recessive traits in a ratio of:
pea plant producing white flowers and introduced pollen
from a purple-flowered plant. He allowed the hybrid
offspring produced by these crosses to self-pollinate for
many generations. Mendel kept a record of the number of
offspring of each type and in each generation.
ta A.
B.
C.
D.
1:1
2:1
2.5:1.5
3:1
on
The hybrid offspring from the experimental crosses are
termed the first filial (Fi) generation. These offspring
always resemble one of their parents. For example, a
cross between wrinkled versus round seeds always
produce Fi progeny with round seeds. Mendel referred to
the Fi trait as being "dominant" over the invisible,
What is the ratio of homozygous to heterozygous
M

"recessive" trait. Mendel allowed individual Fi progeny 8.


plants to self-pollinate, and he observed the offspring in individuals in an F2 generation?
this second filial (F2) generation. He found that some
individuals in this generationexhibited the recessive trait. A. 1:1
r.

B. 3:1
Out of 7,324 F2 individuals resulting from the round- C. 9:3
D. 16:1
seeded Fi self-pollination, 5,474 had round seeds and
1,850 had wrinkled seeds. The ratio was essentially 3:1.
D

Mendel examined all seven traits with alternative


forms and in every case obtained the same ratio results:
3/4 of the F2 individuals exhibited the dominant form of
the trait, and 1/4 displayed the recessive form. He
accounted for his data by forming a central assumption The ploidy number (n) of a cell refers to how many
9.
that alternative forms of a trait are specified by alternative sets of chromosomes are found in its nucleus. For a
alleles, which are discrete. human cell in prophase I, the cell is said to be:
In other words, the factors specified by a pair of A. haploid (n = 1).
alternating alleles are separate. This is known as the law B. diploid (n = 2).
ofallele segregation. C. triploid (n = 3).
D. tetraploid (n = 4).

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Specializing in MCAT Preparation
MCAT- Prep
Biology Gregor Mendel and Inheritance Passage II

10. In an effort to determine whether a purple-flowered


pea plant is heterozygous or homozygous, a test
cross is performed with a white-flowered pea plant.
Based on information in the passage, the MOST
likely conclusion is that the purple-flowered
individual is:

A. heterozygous, if all offspring have purple


flowers.
B. heterozygous, if half of the offspring have
purple flowers.
C homozygous, if all offspring have white
flowers.
D. homozygous, if half of the offspring have
purple flowers and the other half have white

r
flowers.

se
11. The following diagram, illustrating a cross of
wrinkled and round seeded peas, is inaccurate
because the:

P generation

Round
seeds
Wrinkled
seeds
ta
on
M
r.
D

Round seeds Wrinkled seeds

A. P generation is not made up of true breeders.


B. ratio of round to wrinkled seeds in F2 is
incorrect.
C. segregation is incorrectly shown as being
between pods.
D. number of progeny in F2 is incorrect.

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Biology Incomplete Dominance and Codominance Passage III

Passage III (Questions 12-17) 12. Suppose two pink Japanese four-o'clocks are
crossed. What colors would the offspring, the F2
Although many genetic traits are inherited in a generation, exhibit?
classical dominant/recessive pattern, this is not true of
every genetic trait. Certain genes follow the trends of A. 1:2:1, pink:white:red
either codominance or incomplete dominance. B. 1:2:1, red:pink:white
C. All flowers either pink or red
When a heterozygote for a particular trait has a D. 2:2 pink:white
phenotype that is intermediate between those of its two
parents, the genes exhibit incomplete dominance.
Although many examples are known in both plants and
animals, a well-studied example of incomplete dominance
in color occurs in a flower called the Japanese four-
o'clock. When a homozygous white flower (rr) and a 13. Which of the following terms BEST fits this
homozygous red flower (RR) are crossed, the Fi definition?

r
generation is entirely pink (Rr). Both the red color and the
white color are expressed together in the pink offspring. Genes that occupy corresponding loci on
homologous chromosomes and govern variations of

se
the same characteristic.

A. Homologous pairs
Rr Rr B. Pheromones
C. Segregants
Rr Rr

Figure 1. Cross of red and white


ta D. Alleles

flowers, producing all pink offspring.


on
14. Which type of genetic cross is represented in
Another type of genetic expression is codominance, in Figure 1?
which both alleles are expressed independently and
simultaneously in the heterozygote. For example, a horse A. A monohybrid cross.
with a red coat (RR) is crossed with a white horse (rr). B. A test cross.
The foal (Rr) has a mixed red and white coat, in which C. A dihybrid cross.
M

red hairs and white hairs are interspersed. Horse breeders D. An inbreeding cross.
refer to these crossed breeds as being "roan-colored."

A classic example of codominance in humans occurs


r.

in the ABO blood group patterns. A and B refer to two


specific glycoproteins that can be present on the surface
of red blood cells. The A and B glycoproteins are 15. Suppose a gene independent of the one governing
codominant. If either protein is present, it is expressed. the color of a Japanese four-o'clock determined its
D

The lack of either protein results in the O blood type, leaf form in a classical dominant/recessive pattern,
which is recessive to A and B. with straight leaves (L) being dominant over curly
leaves (1). Given the following cross:

RRL1 (Parent 1) X rr Ll (Parent 2)

What is the probability of producinga plant with


pink flowers and curly leaves?

A. 0.25
B. 0.50
C. 0.75
D. 0.00

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Biology Incomplete Dominance and Codominance Passage III

16. The paternity of a royal heir is in dispute. The queen


(known to be the baby's mother) has type AB"
blood, while the king has B\ The baby has A+
blood. The positive or negative Rh factor is
transmitted as a classical Mendelian dominant/
recessive trait, with + being dominant. Is the child
the offspringof the king and the queen?

A. Absolutely
B. Possibly
C. No
D. Unable to tell from this passage

r
17. Two people with type O" blood have a child

se
together. What blood types could their child have?

I. o-
II. A-
III. B"

A.
B.
C.
Ionly
I and II only
I and III only
ta
D. I, II, and III
on
M
r.
D

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