Biology secton test (6)

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BIOLOGICAL SCIENCES TEST 3 TRANSCRIPT
Passage I (Questions 1–4)
1.
The correct answer is choice C. During transcription, DNA acts as a template for the synthesis of a strand of
mRNA. The segment of DNA that acts as the template for transcription is called the anti-sense strand, while the nontemplate strand of DNA, which is identical in sequence to the mRNA being transcribed (except for the presence of
uracil in place of thymine) is called the sense strand. The synthesis of mRNA is catalyzed by the enzyme RNA
polymerase, which moves along the anti-sense strand in the 3' to 5' direction, and synthesizes the mRNA in the 5' to 3'
direction. Free nucleotides pair up with their complementary bases: guanine pairs with cytosine, thymine pairs with
adenine, and adenine pairs with uracil.
So, to answer this question, you need to determine the base sequence of the DNA segment from which this
mRNA was transcribed. You should have immediately eliminated choices A and D because they both contain uracil,
which, as you know, is only found in RNA. To distinguish between choices B and C, you have to follow the basepairing rules and take the polarity of nucleic acid into account. And, if you did so correctly, you should have obtained
the following sequence of bases: CTCAGTCCTTTATCATA in the 5' to 3' direction, which is choice C. If you chose
choice B, it's because you forgot about the polarity. So, choice B is wrong and choice C is the correct answer.
2.
Choice C is the correct answer. This question concerns the process of translation, in which an mRNA
transcript is translated into an amino acid sequence. Translation occurs at the ribosome. Nucleotides are read in
triplets known as codons, and each codon codes for a particular amino acid. The chart of the genetic code matches the
64 mRNA codons with the amino acids to which they correspond. The first letter, or base, of a codon is the left-most
letter, in the 5' to 3' direction; the second base of a codon is in the middle; and the third base is the right-most letter.
As you can see, and as you should know, the genetic code is redundant; that is, an amino acid typically has more than
one codon specifying it. There are also four codons that don't strictly code for amino acids; the codon AUG codes for
the amino acid methionine, but also signals initiation of translation. Therefore, methionine, or Met, is always the first
codon to be translated, and is therefore always the first amino acid in the peptide chain, prior to peptide modification
and processing following translation. During this process, the terminal Met is typically cleaved. There are also three
codons - UAA, UAG, and UGA - that signal termination of peptide synthesis and do NOT code for any amino acid.
All of this is right there on the chart for you, so it's not like you needed to remember all of this stuff to answer the
question.
A ribosome begins translation by reading the mRNA transcript from left to right, in the 5' to 3' direction,
until the start codon - AUG is read. You knew that this transcript must start with the codon AUG because you're told

in the passage that this mRNA fragment contains the first coding nucleotide of v-src. And the first coding nucleotide
will be the first base of the first codon. If you were told that this mRNA fragment did not contain the first codon, then
you would start with the first three bases as the first codon, you wouldn't need to start with AUG. In this particular
sequence, A, U, and G are the second, third and fourth bases, respectively. The second codon consists of the fifth,
sixth, and seventh bases, and so on and so on. Following AUG are the following triplets: AUA, which codes for
isoleucine, AAG, which codes for lysine, GAC, which codes for aspartic acid, and UGA, which is a noncoding codon
signaling the termination of translation. Therefore, the sequence of amino acids in the peptide chain translated from
this strand of mRNA will be: methionine, isoleucine, lysine, and aspartic acid. So choice C is the correct answer. If
you chose choice A, then you probably forgot that the start codon AUG is necessary for initiation. If you chose D,
you probably misread the third codon, which is AAG, as AAC, which codes for asparagine. Choice B is the sequence
obtained if you translated the mRNA beginning with the first three bases, which are UAU, and so on, and then added
on a methionine to the beginning of the sequence. Again, the correct answer is choice C.
3.
The correct answer to this question is choice B. From the question stem you know that you need to find the
experimental observation that most supports the model of cell transformation by v-src. According to the passage, vsrc phosphorylates proteins with its kinase activity, but can only transform cells if it is bound to the plasma
membrane. Now that we've refamiliarized ourselves with the model, let's look at the answer choices. Choice A is
wrong because the substitution of the N-terminal amino acid of v-src, which rendered the cell incapable of kinase
activity, should have prevented v-src from transforming cells. Choice C is wrong for the exact same reason; it doesn't
matter which of the terminal amino acids - the C or the N - is substituted as long as the end result is the same. Choice
B is correct because, assuming that v-src's ability to transform cells is dependent on its ability to bind to
phospholipids, it makes sense that the loss of latter would result in the loss of the former. If the protein reverts back
to the wild type protein, as in choice D, then you would NOT expect v-src capable of cell transformation, since the
wild-type protein is not a transforming protein. How did we know the v-src even has a wild type form? Well, from
the passage, you know that v-src is a transforming protein and that transforming proteins are mutants. And since
mutant proteins must have corresponding wild-type proteins, we know that it's perfectly normal for v-src to have a
corresponding wild-type protein, c-src. Anyway, the fact that the wild type version of v-src does not transform cells
neither supports nor contradicts the model in that v-src must be bound to the plasma membrane and capable of
phosphorylation for it to be capable of transforming cells. Therefore, choice D is wrong. Again, choice B is the
correct answer.

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Kaplan MCAT Biological Sciences Test 3 Transcript

4.
The correct answer is choice D. In a frameshift mutation the reading frame of a mRNA strand is altered
because of the loss or addition of nucleotides, except in multiples of three. Why? Because the remaining nucleotides
are regrouped into a different set of triplets. The protein that is translated as a result of a frameshift mutation is
usually nonfunctional. In this problem, deletion of the fifth base from the 5' end, which is an adenine, changes the
second codon from AUA, which codes for the amino acid isoleucine, to UAA, which is one of the stop codons - a
noncoding triplet that signals the termination of translation. Remember, AUG is the codon that initiates translation.
If you are a little unclear about this point, listen to the explanation to question 2. Thus, our frameshift mutation will
result in the premature termination of translation. With that in mind, let's take a look at the answer choices. Choice
A has the amino acids listed in the sequence as it would have been translated, including the stop codon, but this is
pointless because synthesis halted immediately after the first codon was translated - which is why choice A is wrong
and choice D is the correct answer. If you chose any of the other answer choices, you were probably careless in your
reading of the genetic code chart. Again, the correct answer is choice D.
Passage II (Questions 5–12)
5.
The correct answer is choice B. According to the passage, you know that in a hematocrit, the bottom layer Layer 1- consists of red blood cells; the middle layer - Layer 2 - consists of white blood cells and platelets; and the top
layer - Layer 3- consists of the liquid component of the blood, plasma. If Layer 2 of Patient A's hematocrit is found
to have a greater-than-normal volume, then this must be due to an increase in the number of circulating platelets or an
increase in the number of circulating white blood cells. There are approximately 250,000 - 500,000 platelets per
mm3 of blood, but this volume is fairly static. There are approximately 5,000 - 10,000 white blood cells in
circulation per mm3 of blood; however, this number increases substantially when the body is battling an infection,
since white blood cells are one of the body's main defense mechanisms against foreign invaders such as bacteria and
viruses. During an infection, the body steps up its production of white blood cells and releases more white blood
cells into circulation from its reservoirs. Therefore, choice B is the correct answer, since Layer 2 would be of
greater-than-normal volume in a person with an infection. Choice A, living at high altitudes, WOULD affect the

volume of Layer 1 - the red blood cells. People living at high altitudes tend to have a greater vital capacity, a greater
amount of red blood cells and hemoglobin, and an increased tissue vascularization than people living at lower
altitudes or sea level. All of these adaptations serve to increase the oxygen capacity of the blood, since the
atmospheric partial pressure of oxygen is much lower at high altitudes. Therefore, choice A is wrong. Choice C,
hemophilia, is a sex-linked genetic disease characterized by a deficiency or abnormality of a clotting factor, and would
therefore affect Layer 3, since clotting factors are plasma proteins. Choice D, sickle-cell anemia, would affect the
volume of Layer 1, since this is the layer composed of red blood cells. Again, the correct answer is choice B.
6.
The correct answer is choice A. You're told in the passage that normal hematocrit values range from 40 50% of total blood volume. Looking at Patient B's hematocrit, we see that his red blood cell volume is
approximately 25%, which is way below normal. You're also told in the passage that a person suffering from sicklecell anemia has a low red blood cell count, because sickled red blood cells are very fragile and undergo lysis, thereby
decreasing the volume of circulating red blood cells. So choice A is the correct answer. Looking at the other
answers, we see that choices C and D can be eliminated since both infection and leukemia affect the volume of white
blood cells in the body, and would be reflected by an above-normal volume in Layer 2 of the hematocrit. Choice B,
overproduction of hemoglobin, is one of the consequences of living at a high altitude, as discussed in the explanation
to question 5, and would most likely cause an INCREASE in the volume of Layer 1, not a decrease. Again, the
correct answer is choice A.
7.
The correct answer is choice C. Answering this question relied on your outside knowledge of the structure
of the three different types of blood cells. Red blood cells, which are the only cells found in Layer 1, do NOT contain
nuclei. During the production of red blood cells, which is a process known as erythropoiesis, the nucleus of the
primitive red blood cell shrivels and eventually disappears. Likewise, platelets, which are found in Layer 2, do NOT
contain nuclei; a platelet is a fragment of bone marrow cell containing mitochondria, microtubules, vesicles, and
granules, but no nucleus. Plasma doesn't contain any cells; it consists mostly of solutes dissolved in water, but also
contains respiratory gases, hormones, plasma proteins, waste products of metabolism, and dissolved nutrients.
Therefore, both Layer 1 and Layer 2 contain cells that lack nuclei; and so choice C is the correct answer.
8.
The correct answer is choice D. This is another question that relies on your outside knowledge - this time,
on your knowledge of plasma components. As stated in the passage, the top layer of a hematocrit, which is Layer 3, is
composed of plasma. Plasma consists of water, glucose, amino acids, inorganic ions and salts, fatty acids, respiratory
gases, metabolic wastes, and plasma proteins - of which there are three main types: albumins, fibrinogen, and

globulin. Albumin plays a very important role in the maintenance of blood osmolarity, since albumin is too large to
pass through capillary walls. Therefore, choice D is the right answer. Hemoglobin is the oxygen-carrying molecule
of red blood cells, which are found in Layer 1, so choice A is wrong. Choice B, macrophages, are large white blood
cells filled with lysosomes that engulf and digest foreign material, and hence would be found in Layer 2, which
contains both white blood cells and platelets. Therefore, choice B is also wrong. Likewise, leukocytes, which is just
another way of saying white blood cells, are found in Layer 2, as I just said. Again, choice D is the right answer.

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Kaplan MCAT Biological Sciences Test 3 Transcript

9.
The correct answer is choice C. According to the passage, those individuals who are homozygous for the
HbS allele are afflicted with sickle-cell anemia, and according to the question stem, the disease can be lethal. Yet
despite the high mortality of individuals with the genotype HbS HbS, the frequency of the HbS allele is as high as
40% in some areas of Africa. This is an extremely high frequency for an allele that is essentially lethal in a double
dose. The explanation for this phenomenon lies in the fact that there is a heterozygote advantage; that is, individuals
that have only one copy of the HbS gene have some advantage over normal individuals in areas where malaria is
common. The advantage is that heterozygotes rarely show any of the symptoms of sickle-cell anemia AND are less
likely to die of malaria than normal homozygotes. Why? Because the heterozygote's red blood cells do not
sufficiently support the growth of the malaria parasite. Thus, those individuals possessing a single copy of the HbS
allele have an advantage over those without - an indication that natural selection has favored the maintenance of the
HbS allele frequency in that particular gene pool. The number of heterozygotes that are resistant to malaria and
survive to reproductive age more than outweighs the number of homozygotes that die young of sickle-cell anemia.
Thus, the HbS allele is maintained at a high frequency from generation to generation, and choice C is the correct
answer.
Now let's take a look at the other answer choices. Choice A, mutation, could not ALONE account for the
high frequency of the HbS allele. If mutation were the only evolutionary force at work in maintaining the high
frequency of the HbS allele, this would imply that this mutation arose independently in 40% of certain African

populations, which is highly, highly improbable. Choice B, genetic drift, refers to changes in the gene pool of a small
population due to chance, such as a freak natural disaster that kills everybody in the population that doesn't have a
copy of the HbS allele. It is not mere coincidence that the frequency of the HbS gene is highest in areas where malaria
is common. Choice D is wrong, because reproductive isolation refers to the incompatibility of genitalia, which
prevents interbreeding, typically among different species. Obviously, all human beings are capable of reproducing
with one another, regardless of race, ethnicity, or creed. Again, the correct answer is choice C, natural selection.
10.
The correct answer is choice D. From the passage you know that when the gene coding for the beta-chain for
both HbS and HbA are digested with the enzyme MstII, the HbA gene yields a 1.1-kb fragment because it contains an
MstII site, while the HbS gene yields a 1.3-kb fragment because it lacks an MstII site. Obviously, if you were to take
the 1.1-kb fragment and hybridize it with a probe specific for the 1.1-kb fragment, the hybridization would be
successful. But what about trying to hybridize the 1.3-kb fragment with the same 1.1-kb fragment specific probe?
Well since HbS results from a single amino acid substitution of HbA, this mutation most likely resulted from a point
mutation. A point mutation is the substitution of one nucleotide for another. In other words, if am adenine is present
instead of a thymine at the appropriate location within the gene, HbS will result instead of HbA. This means that the
two genes only differ by one nucleotide and this nucleotide must lie within the MstII site in the HbA gene. So, yes,
the 1.3-kb fragment will successfully hybridize with a probe specific for the 1.1-kb fragment. Why? Because, every
base in the probe that is complementary to the 1.1-kb fragment, except for the one that was substituted, will also be
complementary to the 1.3-kb fragment, and therefore the probe will base pair to both fragments and the hybridization
will be successful.
Now that we understand what is going on, let's look at the question. From the question stem you know that
after digestion of DNA sample with MstII, a 1.3-kb fragment is generated that does successfully hybridize with a
probe specific for the 1.1-kb fragment. This sounds exactly like what happens when the DNA coding for the betachain of HbS is digested by MstII. But does this prove that this DNA sample contains the HbS gene? Well let's look
at the answer choices. Choices A and B are true - the HbS gene does NOT contain an MstII site, and the probe
specific for the 1.1-kb fragment of HbA does bind to the 1.3-kb fragment of the DNA sample. However, this does not
prove that the DNA contains the HbS gene. The only way to prove that the sample contains the HbS gene is to either
match the sequence of the DNA with the known sequence of the HbS gene, or to prove that the DNA does indeed give
rise to HbS protein. Therefore, choices A and B are wrong. O.K., how about C? Choice C is wrong because it's not
true, the probe normally WOULD bind to both fragments. Choice D, however, makes sense. The only reason that the
1.3-kb fragment is generated instead of the 1.1-kb fragment is because the DNA sample lacks an MstII site. But there

is no reason why the only mutation that results in the elimination of this site has to be the mutation that produces the
HbS gene. Actually, any mutation that occurred within the MstII site of the HbA gene would produce a gene without
the MstII site, resulting in the 1.3-kb fragment upon digestion by MstII. And since the only change is in the MstII site,
the probe specific for the 1.1-kb fragment of the HbA gene will still bind to the larger fragment, because almost all of
the bases in the 1.3-kb fragment are still complementary to the probe. Therefore, choice D is the correct answer.
11.
The correct answer is choice D, substitution. HbS hemoglobin is structurally identical to HbA hemoglobin
except for the substitution of a valine for a glutamate at a single point on the molecule. In HbA the codon for
glutamate is GAA; in HbS the codon for valine is GUA. It is evident that in the RNA coding for HbS, adenine has
been replaced by uracil, which means that, working backwards to the DNA from which the mRNA was transcribed,
thymine has been replaced by adenine in the HbS gene. This type of mutation is called a base-pair substitution, which
is when one nucleotide is substituted for another. Thus, choice D is correct. Choice A, deletion, and choice B,
insertion, involve the loss or addition of nucleotides, respectively. Such mutations tend to have very serious effects
on the protein synthesized, if one is synthesized at all, since nucleotides are read as a series of triplets known as

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Kaplan MCAT Biological Sciences Test 3 Transcript

codons. Unless the addition or loss of nucleotides occurs in multiples of three, the reading frame of the RNA will be
altered; this is known as a frame-shift mutation, which is choice C. Again, choice D is the correct answer.
12.
The correct answer is choice B. The probability that the firstborn child of two sickle-cell allele carriers will
be a male carrier is 25%. According to the passage, sickle-cell anemia is a recessively inherited autosomal disorder; it
is NOT X-linked. This means that the gene for hemoglobin is not on the X chromosome, and that the two
characteristics in question - sickle-cell anemia and gender - independently assort during meiosis. Therefore, by
figuring out the probabilities of each independent event and multiplying their individual probabilities, you can
determine the probability that the two events will occur simultaneously. Okay, to figure out the percentage of
offspring that are expected to be carriers in a cross between two carriers, let's do a Punnett square. The genotype is

the same for both parents - HbA HbS times HbA HbS. This is your basic 1:2:1 hybrid cross: 25% of the offspring
will have the genotype HbA HbA and will express the normal phenotype; 50% of the offspring will have the genotype
HbA HbS - and will be carriers of the trait; and 25% of the offspring will have the genotype HbS HbS and will express
the sickle-cell phenotype. So, one half of the offspring will be carriers. Likewise, 50% of the offspring will be male,
and 50% of the offspring will be female, because there is an equal chance of inheriting either the X or Y chromosome
from the father, and a 100% chance of inheriting the X chromosome from the mother. The product of these two
probabilities, 1/2 times 1/2, is 1/4, or 25%. Thus, the correct answer is choice B.
Passage III (Questions 13–17)
13.
The correct answer is choice C. This question asks you to figure out the most likely structure for compound
A based on the information given in the passage. The first classification test is the decolorization of a bromine
solution, which is brownish red. In case you forgot what this test is used for, we'll remind you: it detects double and
triple bonds. What happens is that if a double bond is present, the bromine attacks the bond and attaches itself to it,
forming a dibrominated alkane from a double bond or a tetrabrominated alkane from a triple bond. This uses up the
bromine in the solution, making it turn clear. Now, since all three compounds decolorize the bromine solution, they
must all contain either double or triple bonds. Further on in the passage, it says that compound A also produces a
precipitate when treated with silver nitrate in ammonia, and it yields pentane when treated with excess hydrogen in the
presence of platinum. The silver nitrate test is quite specific and is used to detect triple bonds at the end of molecules.
Compounds with such terminal triple bonds produce precipitates, by a complicated reaction whose details you don't
really need to know. Anyway, since compound A did produce a precipitate, a terminal triple bond must be present
here, and thus choice C is correct. Choice A is incorrect because even though a triple bond is present, it's not at the
terminal position, so the silver nitrate test would be negative. Choice D is incorrect because it has two double bonds,
which would also test negatively. Choice B is incorrect for two reasons. First of all, since it's a ring, this compound
would not form pentane when it is reduced with hydrogen in the presence of platinum; the second is that it also
wouldn't form a precipitate with silver nitrate. Again, the correct answer is choice C.
14.
The correct answer 14 is choice C. This question is similar to question 13 except that this one asks for the
most likely structures for compound C, so you can tackle this question in the same way. We know that Compound C
decolorizes bromine in carbon tetrachloride, so it must be unsaturated, with either double or triple bonds. We also
know that compound C does NOT produce a precipitate when treated with silver nitrate, so it doesn't contain any

terminal triple bonds. Now, to figure out what the possibilities are for this compound's structure, calculate the
hydrogen deficiency of the compound. The formula for a fully saturated hydrocarbon is CnH2n+2, so since compound
C has five carbons, it would have 12 hydrogens if it were fully saturated. Compound C has 8 hydrogens, so it is 4
hydrogens UNDER a fully saturated hydrocarbon. Now, each time a double bond or a ring is formed, two hydrogens
are dropped out of the compound; when a triple bond is formed, four hydrogens are dropped out. Since compound C
has a deficiency of four hydrogens, the compound could contain either a ring and a double bond; two double bonds; or
a triple bond. The last bit of information that we get is that compound C does not form pentane when it is fully
reduced with hydrogen in the presence of a platinum catalyst. Instead, it forms a compound with the formula C5H10,
which has a deficiency of two hydrogens, so we can conclude that compound C contains a ring and a double bond.
Of the choices that we're given, roman numerals I, III, and IV all contain a ring and a double bond, and therefore they
are all possible structures for compound C. Again then, choice C is the correct answer.
15.
The correct answer is choice D. This question asks you to use the information in the passage to pick the
structure which best represents compound B. Let's backtrack a bit to find out what we already know about compound
B. We know compound B is unsaturated, and the fact that it is reduced by hydrogen in the presence of a platinum
catalyst to give pentane shows that the original molecule was not cyclic. Why? Because otherwise, a carbon-carbon
bond would have to be disrupted in order to form a linear molecule like pentane, and hydrogenation won't accomplish
that. Therefore, choice B is incorrect. We also know that compound B doesn't have a terminal triple bond, because it
didn't give a positive result in the silver nitrate test. So, by the same reasoning we used in the last question, the
compound must have either two double bonds or a single, non-terminal triple bond. The answer therefore lies
between choices A, C, and D.
You should know that a strongly basic solution of potassium permanganate will replace a triple bond with a
diketone. If the diketone is then treated with an acidic solution, it will produce two carboxylic acids at the point

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Kaplan MCAT Biological Sciences Test 3 Transcript

between the two keto groups. So, looking over the answer choices, you can see that choice D is the correct response,

since it's the only one that contains a triple bond. Be aware, that a mildly acidic solution of potassium permanganate
oxidatively cleaves double bonds, replacing each of the double bonded carbons with a carboxyl group. Anyway,
knowing this bit of information you can easily eliminate choices A and C. Again the correct choice is D.
16.
The correct answer is choice B. In order to answer this question you need to remember the major uses of
the different types of spectroscopy. Let's briefly review those before we answer the question. First we have infrared
spectroscopy, or simply IR. This is mainly used to determine what functional groups are present in a compound.
Next we have mass spectroscopy, which is generally used to determine a compound's molecular weight. Then we
have nuclear magnetic resonance spectroscopy, or NMR. This is used to determine how the carbons are connected in
the compound--that is, what the carbon skeleton looks like. Finally, there is ultraviolet spectroscopy, which is mainly
used to determine a compound's level of saturation. Okay, back to the question. From the passage, we know that the
molecular weights of all the compounds are the same, so we can't differentiate between them on the basis of their
molecular weight; this rules out mass spectroscopy, and eliminates choices A and C right away. We also know from
the information in the passage that all three compounds are unsaturated, so theoretically we could use UV
spectroscopy, choice D, to differentiate them. However, even though this would tell us whether they had double or
triple bonds, it wouldn't give us the exact structure. For instance, it wouldn't reveal whether the multiple bonds were
terminal or non-terminal. Notice that the question asks which of the choices would be MOST useful; this means that
even though you've found a possible answer, you should still check to see if there's another choice that's better.
Choice B, an IR in conjunction with an NMR, would be the most useful possible set of instruments, because it would
reveal both the functional groups in the compounds AND the shape of the carbon skeleton. Therefore, choice B is the
best and correct answer.
17.
The correct answer is choice C. This is a negative response question in that you have to pick the answer
choice which would NOT be a possible structure for the compound with the formula C8H12O2. Well, if you draw out
each answer choice it should be pretty evident that choice C will not have the formula C8H12O2. Choice C is a
straight chain carbon molecule, so to begin with, you need to draw out 8 carbons. Then you have to draw in the
aldehyde functionality at one end of the molecule. You then need to draw in three carbon double bonds and a
hydroxyl group. Whether you draw the double bonds isolated, conjugated or cumulated, you should be able to see
that the maximum number of hydrogens in the molecule is 10, and so the overall formula for this compound is
C8H10O2, making it the correct answer.

If you do the same sort of thing with answer choice A, but this time draw in a non-terminal triple bond along
with a hydroxyl functionality and an aldehyde group, you can see that this molecule WILL have the formula C8H12O2
and so it is incorrect. Choice B will also have the formula C8H12O2, even if you draw the carbonyl groups as two
aldehydes, two ketones or one ketone and one aldehyde functionality. Finally, choice D is also incorrect. This one is
a bit more tricky, as the straight chain will consist of 5 carbons, due to the attachment of a cyclopropane group to one
of the carbons. Also interrupting this 5 carbon stretch is an ester linkage. You should be able to see that any
combination of the functional groups--whether they are drawn as terminal or non-terminal--will lead to the formula
C8H12O2. Again then, choice C is the correct answer.
Discrete questions
18.
The correct answer is choice A. Nondisjunction is defined as the failure of homologous chromosomes to
properly separate during meiosis I, OR the failure of sister chromatids to properly separate during meiosis II. The end
result is the production of abnormal gametes; half of the gametes wind up with two copies of the same chromosome,
the other half wind up with no copies of that chromosome. If one of these gametes fertilizes a normal gamete, the
resulting zygote will either have three copies of that chromosome, which is called trisomy, or one copy of that
chromosome, which is called monosomy. Since nondisjunction results in the movement of an entire, intact
chromosome, the base sequence of the nondisjunctioned chromosome remains unchanged; so, choice A is the right
answer. Most monosomies and trisomies are lethal, causing spontaneous abortion during pregnancy. One of the most
common trisomies seen in humans is Down's syndrome, which results from the nondisjunction of chromosome 21.
Choice B, base substitution, is exactly what its name implies, a substitution of one base for another, which
obviously results in a change in the base sequence of DNA. Therefore choice B is wrong. Choice C, translocation, is
when a chromosome breaks, and the fragment that breaks off joins with a nonhomologous chromosome. This means
that a new piece of DNA is added to a pre-existing chromosome, causing a change in the base sequence; therefore,
choice C is also wrong. Choice D, recombination, is by definition, the creation of new gene combinations by way of
sexual reproduction and crossing over in eukaryotic organisms, and by way of transduction, transformation, and
conjugation is prokaryotic organisms. New gene arrangements necessarily implies that there is a change in base
sequence, and so choice D is also wrong. Again, choice A is the correct answer.
19.
The correct answer is choice D. There are two reasons why the single bond length is shorter than expected in
1,3-butadiene. First, the pi bonds that overlap and consequently form the two double bonds in 1,3-butadiene also

‘spill over’ into the single bond. This results in partial double bond character and a shorter bond length. Therefore,
statement I is correct.

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Kaplan MCAT Biological Sciences Test 3 Transcript

Orbital hybridization also goes some way to explaining why the bond length is shorter. In the ethane
molecule, and indeed any alkane, the carbon-carbon bonds are formed by the overlap of sp 3 orbitals. However, in 1,3butadiene, the carbon-carbon single bond is formed by the overlap of sp 2 orbitals. As a result, there is 33% s
character in this bond, not 25% s character like in the sp 3 bond. This increase results in a shorter and stronger bond
and so statement III is also correct. Therefore, choice A is incorrect and again, D is the right answer.
Statement II is correct if you are describing an observable trend in the table. However, it is incorrect if you
are using it to reason why the 1,3-butadiene single bond length is shorter. There are a greater number of carbons in
1,3-butadiene than in ethane and propene, but this does not account for the shorter bond length. Therefore, choices B
and C are wrong and choice D is the right answer.
20.
The correct answer is choice B. This question is pure outside knowledge; there really isn't any reasoning
involved at all. The pancreas is both an endocrine and an exocrine organ. As an endocrine organ, the pancreas
secretes the hormones glucagon, insulin, and somatostatin. As an exocrine organ, the pancreas secretes the enzymes
trypsin, choice C, carboxypeptidase, choice D, lipase, choice A, and chymotrypsin, which is not one of our choices,
into the small intestine. Chymotrypsin, trypsin, and carboxypeptidase are actually secreted by the pancreas in their
inactive forms, chymotrypsinogen, trypsinogen, and procarboxypeptidase, respectively, and are converted into their
active forms in the small intestine. Chymotrypsin and trypsin hydrolyze specific peptide bonds; carboxypeptidase
hydrolyzes terminal peptide bonds at the carboxyl end; and lipase hydrolyzes lipids. Therefore, choices A, C, and D
are wrong. On the other hand, aminopeptidase, choice B, is an enzyme secreted into the small intestine by the
intestinal glands. Aminopeptidase is also involved in protein digestion; it hydrolyzes terminal peptide bonds at the
amino end. Therefore, choice B is the correct answer.
21.
The correct answer is choice B. This is another organic chemistry question, which asks about the number of

acidic hydrogens in the four compounds listed. Choice A is a diacid with two methoxy groups attached to the parent
chain. The two methoxy groups are on the two carbons adjacent to the carboxyl groups. Since these methoxy groups
are electron-donating, they destabilize the carboxylate anions, so these carboxyl groups have pretty low acidity. Since
choice A doesn't contain any very acidic hydrogens therefore, you can eliminate it. Choice B is another dicarboxylic
acid--but it has an electron-withdrawing sulfonate substituent adjacent to both carboxyl groups. This will stabilize
both carboxyl anions making both carboxyl hydrogens highly acidic. In addition, the hydrogen on the central carbon is
also very acidic, because it's surrounded by the three electron-withdrawing groups. Therefore, choice B contains three
highly acidic hydrogens. Choice C contains just one highly acidic hydrogen--the one on the carboxyl group. Finally,
choice D also contains 1 acidic hydrogen. The hydroxyl proton in phenol is weakly acidic since the benzene ring can
stabilize the negative charge of the phenoxide ion. However, choice B still has the greatest number of acidic
hydrogens, making it the correct answer.
22.
Choice A is the right answer. This question tests your knowledge of eukaryotic, prokaryotic, and viral
structure. Viruses are composed of a nucleic acid encapsulated by a protein coat - period. Bacteriophage are viruses
that infect bacteria only. Since the answer to the question is an organism characterized by the presence of a cell wall,
you can rule out choice B, bacteriophage, and choice C, viruses, which leaves you with only two choices. Choice D,
fungi, are eukaryotic, heterotrophic organisms, and as such, have membrane-bound organelles. Fungi do have cell
walls, which are composed of chitin, but they also have nuclei bound by nuclear membranes, so choice D is also
wrong. Which leaves us with choice A, bacteria. Bacteria are members of the kingdom Monera - a group of
unicellular, primitive, prokaryotic organisms. Prokaryotes are characterized by a lack of membrane-bound organelles.
Bacteria typically have a cell wall, a cell membrane, ribosomes, and a circular chromosome located in a region of the
cell known as the nucleoid. Therefore, of the answer choices, it is the bacteria that is characterized by the presence of
a cell wall but the lack of a nuclear membrane. Again, choice A is the right answer.

Passage IV (Questions 23–27)
23.
Choice D is the correct answer. If you look at Figure 1 you'll see that those athletes on a high protein and
high fat diet did not recover to pre-exercise glycogen levels, EVEN after 5 days of recovery. The initial level of
muscle glycogen was about 15 g/kg, and the level after 5 days was about 9 g/kg. This seems to indicate that eating a
high protein and fat diet incurs a great loss of glycogen. Think about this for a minute. The body needs ATP, and the

majority of this ATP is generated from the catabolism of glucose during cellular respiration. In the absence of
glucose, the body draws on its other nutrient stores to compensate for this absence. Proteins and fats are degraded
into molecules that can enter the aerobic respiration pathway, and the stores of glycogen in the liver and muscle cells
are converted back into glucose. And this is exactly what happens to a person on the high protein weight loss diet
described in the question; by depriving the body of carbohydrates, the stores of glycogen are converted into glucose,
which causes a substantial weight loss early on in the diet. Thus, choice D is the right answer. Choice A is wrong
because there is no reason to conclude that there is an increase in metabolic activity on a high-protein diet. Choice B
just doesn't make much sense; mammals are homeotherms, which means that a constant body temperature is

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Kaplan MCAT Biological Sciences Test 3 Transcript

maintained. If there is an increase in body temperature not caused by fever, thermoregulatory mechanisms go to work
to return internal temperature to normal. Therefore, choice B is wrong. Choice C is wrong because there is not a
substantial loss of fat on an all protein diet; loss of fat is just wishful thinking on the part of the dieter. Again, choice
D is the correct answer.
24.
The correct answer is choice C. You've got to determine which of the energy systems are used to supply the
energy needed to sustain physical activity for 22.3 seconds, which is the amount of time it takes a world-class
swimmer to swim 50 meters in the freestyle event. You don't have to know anything about swimming to answer the
question because the only thing that you need to know, which is the amount of time the activity takes, is given to you
in the question stem. And according to the passage, the phosphagen system can sustain up to 15 seconds of maximal
activity. But this energy system alone isn't enough to complete the 50-meter freestyle, so choice A is wrong. Since
the swim takes longer than 15 seconds, you know that the glycogen-lactic acid system must also be involved. The
glycogen-lactic acid system provides energy for 30 to 40 seconds in addition to the phosphagen system's 15 seconds.
Therefore the glycogen-lactic acid system must take over where the phosphagen system left off to enable the swimmer
to complete the event. Therefore, choice C is correct. Since all physical activity begins with the phosphagen system,
choice B is incorrect. The aerobic system is not used for swimming a distance as short as 50 meters in 22.3 seconds,

so choice D is wrong. Again, choice C is the correct answer.
25.
Choice D is the correct answer. After three hours of exercise, let alone two, it can be assumed that the
phosphagen system of those athletes on the high protein and fat diet is completely used up, since this system can
support activity for a maximum of 15 seconds, according to the passage. So, choice A is wrong. And, if you look at
the graph, you'll see that after two hours of exercise, those athletes on the high protein and fat diet have completely
depleted their muscle glycogen stores. And if they exercise for yet a third hour, there's no recovery period during
which the muscle glycogen supply can be replenished. Well, this means that the glycogen-lactic acid system cannot
provide any of the energy needed for the third hour of exercise, because the glycogen-lactic acid system supplies
energy by breaking down muscle glycogen into lactic acid. No more glycogen, no more glycogen-lactic acid system.
This means that you can eliminate choices B and C, because they both contain the glycogen-lactic acid system. The
only remaining energy system is the aerobic system, which is the only energy system capable of sustaining the body's
energy needs during extended periods of strenuous exercise. Therefore, choice D is the correct answer.
26.
The correct answer is choice D. From the question stem you know that you need to find which of caffeine's
effects would be deleterious to an athlete. First of all, you should realize that you do not need to know anything
about caffeine to answer this question. All you have to do is evaluate the list of effects in the answer choices. So let's
look at the choices. Is increasing the concentration of epinephrine, as in choice A, a good or bad thing? Epinephrine
increases the conversion of glycogen to glucose in the liver and muscle tissue, causing a rise in blood glucose
concentration and an increase in basal metabolic rate. Well, if more glucose is made available to the muscle, the
muscle will be able to generate more energy to perform the athletic activity. Thus, this enhances the athlete's
performance, and so choice A is incorrect. What about systemic vasodilation, as in choice B? Vasodilation of the
systemic system will allow more blood to flow into the organs and muscle tissue of the body, thereby providing an
increased oxygen supply to the athlete’s' muscles. Increasing the oxygen supply will increase the rate of cellular
respiration, thereby generating more energy for the muscle. So, this too is beneficial to the athlete, and thus choice B
is wrong. In humans, this vasodilation is short-lived, so this effect would only benefit athletes in the beginning of an
event. What about relaxing the muscles of the bronchi, as in choice C? The bronchi are the pair of respiratory tubes
branching into either lung at the lower end of the trachea. They subdivide into progressively finer passageways, the
bronchioles, culminating in the alveoli. The alveoli are the sites of gas exchange between the air in the lungs and the
blood in the capillaries. So by relaxing the bronchi muscles, more air is allowed to enter the lungs and thereby

increase the partial pressure of oxygen in the blood. As a result, more oxygen will be delivered by the blood to the
muscles. This increased oxygen supply will increase the rate of energy production. Therefore, this effect of caffeine
is also helpful to the athlete, so choice C is also incorrect. So by the process of elimination, choice D is the correct
answer. ADH stimulates water reabsorption in the kidneys. So if ADH secretion is inhibited by a diuretic such as
caffeine, water reabsorption will decrease and dilute urine will be collected in the bladder. If water reabsorption does
not return to normal, dehydration can result. Since dehydration would adversely affect the athlete's performance,
choice D is the correct answer.
27.
The correct answer is choice A. As you're told in the passage, the glycogen-lactic acid system converts the
muscle's stores of glycogen into glucose, which is then converted into lactic acid in the absence of oxygen. The lactic
acid diffuses out of the muscle cells and into the bloodstream and interstitial fluid. During the recovery period, this
lactic acid must be removed. This is accomplished in two ways. First, some of the lactic acid is converted into
pyruvate, which is metabolized by the Krebs cycle. Second, most of the lactic acid is used to replenish the depleted
glycogen stores in muscle cells. This is done by first converting the lactic acid into glucose in the liver, and, in turn,
this glucose is converted into glycogen. And, if you look at the graph, you see that no matter what type of diet the
athletes were on, it's obvious that the body tries to replenish its glycogen stores during the recovery period following
exercise. Therefore, choice A is the right answer. You could also have gotten the right answer by the process of
elimination, so let's take a look at the other choices for a moment. I hope that you weren't fooled by choice B.

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Kaplan MCAT Biological Sciences Test 3 Transcript

Although lactic acid might sound a little bit like lactose, it has nothing to do with milk, or calcium, and is not stored
in bone. Choice C is wrong, because as already said, some of the lactic acid will be reconverted to pyruvate acid in
the liver and fed into the Krebs cycle, which is a very long way from being stored as fat. Choice D is wrong because
in the glycolytic pathway, glucose is degraded into pyruvate, and in the absence of oxygen, pyruvate is converted into
lactic acid. The body's "goal" is to get rid of the lactic acid that it produced during the exercise, not create more of it.
Again, choice A is the correct answer.

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