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MCAT Topical Tests

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BIOLOGY TOPICAL:

Generalized Eukaryotic CellTest 1

Time: 21 Minutes*Number of Questions: 16

* The timing restrictions for the science topical tests are optional. Ifyou are using this test for the sole purpose of contentreinforcement, you may want to disregard the time limit.



MCAT

2 as developed by

DIRECTIONS: Most of the questions in the followingtest are organized into groups, with a descriptivepassage preceding each group of questions. Study thepassage, then select the single best answer to eachquestion in the group. Some of the questions are notbased on a descriptive passage; you must also select the

best answer to these questions. If you are unsure of thebest answer, eliminate the choices that you know areincorrect, then select an answer from the choices thatremain. Indicate your selection by blackening thecorresponding circle on your answer sheet. A periodictable is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS

1

H

1.0

2

He

4.0

3Li

6.9

4Be

9.0

5B

10.8

6C

12.0

7N

14.0

8O

16.0

9F

19.0

10Ne

20.2

11

Na

23.0

12

Mg

24.3

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

Cl

35.5

18

Ar

39.9

19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9

24

Cr

52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8

37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag

107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3

55

Cs

132.9

56

Ba

137.3

57

La *

138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg

200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222)

87

Fr

(223)

88

Ra

226.0

89

Ac †

227.0

104

Rf

(261)

105

Ha

(262)

106

Unh

(263)

107

Uns

(262)

108

Uno

(265)

109

Une

(267)

*

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

Lu

175.0

†

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np

(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

Cf

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)



Generalized Eukaryotic Cell Test 1

KAPLAN 3

Passage I (Questions 1–6)

It has been proposed that the nervous tissue of theelectric eel, Electrophorus electricus, may be useful in therepair of damaged nerves in humans. To isolate thevarious subcellular structures within the nerve cells of Electrophorus electricus, a researcher uses rate-zonalcentrifugation. This technique separates the various

subcellular structures by size and density. Aftercentrifugation, the large dense structures typically form apellet at the bottom of the test tube, while the lighter,less dense structures remain suspended in the supernatant(the fluid layer). By centrifuging at different speeds anddurations, different organelles can be isolated.

The eel nerve cells are suspended in a 0.25 Msucrose solution, which is isotonic to the cells. Most of the cells are broken open by stirring this suspension in ahigh-speed blender to form a cell homogenate. Thehomogenate is then filtered to remove any unbroken cells.The filtered homogenate is centrifuged at increasing speedsfour times, with a pellet collected after each step, and the

supernatant transferred to each successive tube. Thisprocedure is outlined in Figure 1.

Centrifuge

600g x

10 min

Pour out:

15,000g x5 min

Pour out:

100,000g x60 min

Pour out:

300,000g x2 hr

Filteredhomogenate

Pellet 1 Pellet 2 Pellet 3 Pellet 4

Supernatant

4

Supernatant

3

Supernatant

2

Supernatant

1

Figure 1

1 . Analysis of Pellet 2 showed that it possessed highsuccinate cytochrome C reductase activity. Thisenzyme oxidizes succinic acid to fumaric acid, bothintermediates of the Krebs cycle. Therefore, this pelletmost likely contains:

A . mitochondria.B . plasma membrane.

C . endoplasmic reticulum.D . cytoplasm.

2 . To determine the activity of the enzyme lactatedehydrogenase, the compound DPNH is added to eachsample of supernatant. For every molecule of pyruvate that is reduced to lactate by lactate

dehydrogenase, a molecule of DPNH is oxidized:

Pyruvate + DPNH→ Lactate + DPN

In which of the following samples would you expectlactate dehydrogenase activity to be found?

I. Supernatant 1II. Supernatant 2

III. Supernatant 3IV. Supernatant 4

A . I onlyB . III and IV onlyC . II, III, and IV only

D . I, II, III, and IV

GO ON TO THE NEXT PAGE.



MCAT

4 as developed by

3 . Which of the following molecules would you expectto find in the pellet containing the plama membrane?

-O

NH

N

N

O

NHN

O

OH

HH

HHOH

OP

O

O-

I.

3

3

P

H2C

HC

CH2

O C

O

(CH)16 CH

O C (CH)16

O

CH

O

O O-

O-II.

P-OO

H

HH

HH

N

N

NH2

O

OH

O

O

O-

III.

A . I onlyB . II onlyC . III onlyD . I and III only

4 . Which of the following structures would most likelybe found in a higher concentration in the plasmamembrane of eel nerve cells as compared to theplasma membrane of eel epithelial cells?

A . GlycoproteinsB . CholesterolC . Acetylcholine receptors

D . Phospholipids





KAPLAN 5

5 . The enzyme DNase is added to Pellet 1. After thisenzymatic digestion, which of the following wouldmost likely be found within the digested pellet?

A . mRNAB . HistonesC . RhodopsinD . Elongation Factor 2

6 . The researcher believes that Pellet 4 containsribosomal subunits. Which of the followingobservations would most support this hypothesis?

A . Pellet 4 was found to contain uncharged tRNA.B . Pellet 4 was found to contain two molecules

with sedimentation coefficients of 40S and 60S.C . Pellet 4 was found to contain the enzyme

peptidyl transferase.D . Supernatant 4 was found to contain unprocessed

proteins.

Passage II (Questions 7–12)

The yeast Schizosaccharomyces pombe divides byfission and is thus an ideal microorganism for the study of cell division in higher eukaryotic cells. A scientistbelieves that she has identified the region of the S. pombechromosome containing the DNA sequence that includesthe centromere. From this region she has isolated two

different fragments of DNA: Fragments I and II. Thecentromere is the region of the chromosome thatmaintains the attachment of sister chromatids duringmeiosis I and includes the site of attachment to themicrotubules of the meiotic and mitotic spindles. Duringdivision, the spindle must both assemble and disassembleits tubulin subunits to properly function in thesegregation of chromosomes.

To test which of the two fragments contains thecentromere region, the scientist takes a mutant strain of S . pombe cells that cannot grow unless adenine is providedin the growth medium, and performs the following set of experiments.

Experiment 1Mutant cells are inoculated onto an agar plate

containing minimal medium.

Experiment 2The normal gene for adenine biosynthesis (adenine+)

is inserted into a plasmid, which is then inserted into themutant cells. The cells are then inoculated onto agar platescontaining minimal medium. (A plasmid is a self-replicating extrachromosomal DNA fragment.)

Experiments 3A and 3BFragment I and Fragment II are inserted into separate

plasmids containing the adenine+ gene. The plasmids arethen inserted into the mutant cells, which are inoculatedonto separate agar plates containing minimal medium. If the plasmid contains the DNA fragment with thefunctional centromere, it will be stably inherited bydaughter cells during cell division.

The experimental results are shown in Table 1.




MCAT

6 as developed by

Table 1

Expt. Mutant S. pombeon minimal

medium with:

Increase incell # after1st round

of division

Visiblecoloniesafter 20

rounds of division

1 cells alone No No

2* cells + adenine+ Yes No3A* cells + adenine+ +

Fragment IYes No

3B* cells + adenine+ +Fragment II

Yes Yes

*Cells with introduced plasmid

7 . Based on the information in the passage, it can beconcluded that:

A . Fragment I contains a centromere.

B . Fragment II contains a centromere.C . Fragments I and II both contain centromeres.D . the mutant S. pombe chromosomes do not

contain centromeres.

8 . The results obtained in Experiment 3B can be changedsuch that they are identical to the results obtained inExperiment I if:

I. a drug that inhibits microtubule assembly isadded to the growth medium used in

Experiment 3B.II. a drug that inhibits microtubule disassembly is

added to the growth medium used inExperiment 3B.

III. a drug that inhibits cytokinesis is added to thegrowth medium used in Experiment 3B.

A . I onlyB . II onlyC . I and II onlyD . I, II, and III

9 . Minimal medium contains only salts and sugar. Theprimary reason that the cells used in Experiment 1cannot grow on minimal medium is that these cellsare NOT able to synthesize:

A . protein.B . essential fatty acids.C . nucleic acids.

D . salts.

1 0 . Which of the following graphs best represents thefour growth curves obtained under the experimentalconditions summarized in Table 1?

B .

C .

D .

A .

l o g c

e l l #

# of cell divisions

# of cell divisions

# of cell divisions

l o g c

e l l #

l o g c

e l l #

l o g c

e

l l #

# of cell divisions1 20 1 201 20

1 20 1 20

Key: Expt 1 Expt 2

Expt 3AExpt 3B





KAPLAN 7

1 1 . The cells used in Experiment 2 are most likely ableto increase their cell number after one round of celldivision because:

A . the replicated adenine+ plasmid binds tomicrotubules and segregates to daughter cellsduring mitosis.

B . the replicated adenine+ plasmid binds to

microtubules and segregates to daughter cellsduring meiosis.

C . the replicated adenine+ plasmid binds tomicrofilaments and segregates to daughter cellsduring cell division.

D . the replicated adenine+ plasmid randomlysegregates to daughter cells during cell division.

Questions 12 through 16 areNOT based on a descriptivepassage.

1 2 . In vitro, the transcription factor SP1 binds nucleicacids with a high affinity. Radio-labeled SP1 wouldmost likely be found in all of the following organellepreparations EXCEPT:

A . nucleoli.B . mitochondria.C . ribosomes.D . Golgi apparati.

1 3 . All of the following are involved in directing proteinsto their final destination within a cell EXCEPT:

A . amino acid sequence of the protein.B . vesicles.C . lysosomes.D . endoplasmic reticulum.

1 4 . A researcher discovered that in order for the HAP2protein to elder the nucleus and bind to the CYC1gene, a molecule of ATP must be consumed. Whichof the following transport mechanisms must beinvolved in the movement of HAP2 into the nucleus?

A . Active transportB . Facilitated diffusionC . Passive diffusion through the nuclear poresD . Endocytosis




MCAT

8 as developed by

1 5 . Three funnels, each containing a differentconcentration of a urea solution, are covered with asemi-permeable membrane that is impermeable tourea. The funnels are inverted and placed in a beakercontaining a different urea solution What is theconcentration of the solution?

A . 2.0 M ureaB .

1.5 M ureaC . 1.0 M ureaD . 0.0 M urea

dH201Murea

2Murea

Urea Solution

Initial state

dH201Murea

2Murea

Urea Solution

Final state

1 6 . Which of the following would most likely play a rolein cellular adhesion and recognition?

A . Carbohydrate molecules on the exterior of theplasma membrane

B . The fluidity of the plasma membraneC . Peripheral proteins on the interior of the plasma

membraneD . The hydrophobic regions of the plasma

membrane

END OF TEST




KAPLAN 9

ANSWER KEY:

1. A 6. B 11. D 16. A

2. D 7. B 12. D

3. B 8. D 13. C

4. C 9. C 14. A

5. B 10. A 15. C



MCAT

10 as developed by

GENERALIZED EUKARYOTIC CELL TEST 1 EXPLANATIONS

Passage I (Questions 1-6)

1. The correct answer is choice A. Although the question stem may have scared you a little bit, this question is really

quite easy. The only thing you needed to get out of the question stem is that Pellet 2 was found to have one of the enzymesof the Krebs cycle. So as you can see, this question is simply asking you where the Krebs cycle occurs. Well fromintroductory biology, you should know that the Krebs cycle occurs in the mitochondria, along with electron transport andoxidative phosphorylation. Therefore choice A is the correct answer. Just to review briefly, in eukaryotic cells, glucose isbroken down to pyruvate in the cytoplasm in a series of reactions known as glycolysis. The pyruvate then enters themitochondria, where it is converted into acetyl CoA. The acetyl CoA then enters the Krebs cycle. All of the NADH andFADH2 molecules generated during these processes are shuttled into the electron transport chain, which is coupled with

oxidative phosphorylation to produce ATP.Anyway, back to the question. As we just learned, glycolysis, not the Krebs cycle, occurs in the cytoplasm, and so

choice D is incorrect. The other two choices have nothing to do with the Krebs cycle at all. The plamsa membrane surroundsthe cell and acts as a barrier between the cell and the outside environment and regulates the passage of material into and out of the cell. So choice B is incorrect. Finally, the endoplasmic reticulum, choice C, is involved with protein processing and isthe site of ribosomal attachment during translation. Thus choice C is also wrong.

2. Choice D is the correct answer. In order to answer this question you need to understand the basic principle of centrifugation as explained in the passage, as well as the information provided in the question stem. From the question stemyou know that DPNH is used to monitor the conversion of pyruvate to lactate. Although you have probably never heard of DPNH, you should recognize the reaction of pyruvate going to lactate as the fermentation step that occurs in most eukaryoticcells under anaerobic conditions. Well, where in the cell does fermentation occur? Glycolysis occurs in the cytoplasm, whilethe Krebs cycle and electron transport/oxidative phosphorylation occur in the mitochondria.

So now that we’ve deciphered the question stem, let’s see how centrifugation works. From the passage you knowthat larger objects will form a pellet at the bottom of a tube, while lighter, less dense structures remain suspended in thesupernatant. You also are told in the passage and shown in Figure 1, that the supernatant from the first tube is transferred tothe second, and so on and so on. Therefore, cytoplasm will be present in ALL four supernatants and so choice D is the correctanswer.

3. The correct answer is choice B. The question is essentially asking, “the plasma membrane is made up of what

molecules?” The plasma membrane is composed of phospholipids, a molecule that has a glyerol backbone attached to twofatty acids and a phosphate group. The phospholipids are arranged in a bilayer, which is why the cell membrane is alsoreferred to as the “phospholipid bilayer.” Roman numeral II is a phospholipid, so must appear in the correct answer. Answerchoice C is the only one that includes Roman numeral II. Molecule I is an RNA molecule, and III is a DNA molecule.Neither RNA or DNA make up the plasma membrane. Another possible molecule could be a protein, as the plasmamembrane has receptors and ion channels/pumps studded throughout its length.

4. Choice C is the correct answer. You’re being asked to deduce what will be found in a higher concentration in theplasma membrane of an eel nerve cell as compared to the membrane of an eel epithelial cell. Well, what are the normalcomponents of a generic plamsa membrane? A eukaryotic plasma membrane consists of a phospholipid bilayer. Sophospholipids are a common element of all plasma membranes and therefore would not be found in a higher concentration innerve cell membranes. Thus choice D is incorrect. Cholesterol molecules are embedded in the hydrophobic interior of thebilayer. Therefore, cholesterol is also a common component of plasma membranes. So choice B is incorrect. Glycoproteins,which are proteins that contain carbohydrate components, extend out of the plasma membrane and function in cell adhesion

and recognition. Therefore, glycoproteins are also common plasma membrane elements, and thus choice A is incorrect. So bythe process of elimination, choice C is the correct answer. if you think about it, this answer makes sense. Acetylcholine isone of the principal neurotransmitters responsible for the transmission of a nerve impulse, and its message is transmittedwhen it binds to acetylcholine receptors found on the plasma membrane of its target cells. Since acetylcholine is aneurotransmitter, the plasma membrane of an epithelial cell would not be expected to have a high concentration of acetylcholine receptors. But you would most likely find a high concentration of acetylcholine receptors on the plasmamembrane of eel nerve cells. Therefore, choice C is the correct answer.

5. Choice B is the correct answer. Although you are not explicitly told what is in Pellet 1, you should have been ableto figure it out. First of all, you’re told that DNase is added to Pellet 1. Well, you know that the only place DNA is found in




KAPLAN 11

an animal cell is within the nucleus and the mitochondria. And according to the passage, larger, denser organelles will pelletout first. Since the nucleus is larger than any mitochondrion, it makes sense that the nucleus would probably be in Pellet 1,while mitochondria would remain in supernatant 1. So now we know that Pellet 1 contains eel nerve cell nuclei. What effectwill DNase have on the nucleus? DNase is an enzyme that degrades DNA by attacking the phosphodiester bonds that hold ittogether. So adding DNase to the nucleus will destroy the DNA of eukaryotic chromosomes, leaving behind the DNA bindingproteins that are an integral part of eukaryotic chromosome structure. These DNA binding proteins are known as histones;there are five major types of histones. Keep in mind that prokaryotes lack histones. Therefore choice B is correct. Choice A iswrong because a DNase only degrades DNA, and therefore mRNA would not be an end product of its action. RNases degradeRNA. Rhodopsin, choice C, is a protein found in the external segments of the rods of the retina. Like most proteins,rhodopsin is not located within the nucleus of retinal cells, nor in that of nerve cells. Furthermore, DNase activity would notyield rhodopsin. Therefore, choice C is incorrect. Choice D is also incorrect for the same reason. Elongation Factor 2 is aprotein that associates with ribosomes during the addition of each amino acid during translation.

6. Choice B is the correct answer. Since eukaryotic cells were used, the researcher must believe that Pellet 4 containseukaryotic ribosomal subunits. But before we look at the answer choices, let’s briefly review the properties of eukaryoticribosomes. Eukaryotic ribosomes consist of a 60S subunit and a 40S subunit, which come together to form an 80S complex.Both of the subunits are comprised of proteins and rRNA. Prokaryotic ribosomes, on the other hand, consist of a 30S subunitcombined with a 50S subunit, forming a 70S complex.

Now let’s look at the answer choices. Uncharged tRNA, choice A, is typically found in the cytoplasm. UnchargedtRNA, which is tRNA without an amino acid, is only transiently associated with the whole ribosome after it has transferredits amino acid to the growing peptide chain. Nor would uncharged tRNAs be associated with the individual ribosomalsubunits, except for the initiator methionine tRNA, which does bind to the 40S subunit. Therefore, choice A definitely doesNOT support the researcher’s hypothesis of ribosomal subunits in Pellet 4. Thus, choice A is incorrect. The presence of a40S and a 60S molecule in the pellet would support the presence of the ribosomal subunits. Therefore this does support theresearcher’s hypothesis. And although it DOES support the hypothesis, it in no way PROVES that Pellet 4 containsribosomal subunits. But, so far, this seems like a pretty good answer. Let’s check out the remaining two choices. Theenzyme peptidyl transferase is responsible for the formation of peptide bonds during translation. And like the uncharged tRNAmolecules, peptidyl transferase is never bound to the ribosome itself, and would only be found in association with the wholeribosome, not its individual subunits. Therefore choice C is also incorrect. The contents of the supernatant associated withPellet 4 does not give us any information about Pellet 4 itself, so it neither supports nor contradicts the notion that Pellet 4contains ribosomal subunits. So choice D is also incorrect. Therefore, choice B must be the correct answer.

Passage II (Questions 7-11)

7. The correct answer is choice B. To answer this question you must utilize information contained in the passage andthe data in Table 1. We see that in Experiment 1, the mutant cells cultured in minimal medium cannot grow. And the reasonwhy the mutant strain is unable to grow on minimal medium is that it lacks adenine. Remember, the passage specificallystates that the mutant strain CANNOT grow in the absence of adenine. Although the passage does not explicitly state that theminimal medium used in the experiments lacks adenine, it is the most obvious conclusion that can be drawn given that themutant strain doesn’t grow in it. Minimal medium contains only the bare essentials for a wild-type organism to grow, andsince wild-type organisms are able to synthesize adenine on their own, there is no reason for minimal medium to besupplemented with adenine.

In Experiments 2, 3A, and 3B, the mutant yeast cells that were inserted with plasmids containing the gene that codesfor adenine were able to complete at least one round of cell division on minimal medium, according to Table 1. Therefore wecan conclude that the mutant yeast cells CAN grow on minimal medium if they contain a plasmid with the adenine+ gene.The passage also implies that if a plasmid does not have a centromere, the plasmid will not be stably maintained in the

mutant yeast cell colony. And if a cell doesn’t have the plasmid, it will die, since it can’t synthesize adenine on its own. Onlythose cells with plasmids containing centromeric DNA and the adenine+ gene will be able to form visible colonies.Now that we understand the logic behind the data presented in Table 1, let’s look at the answer choices. From the

table you see that in Experiment 3A, the yeast with Fragment I produced no visible colonies after 20 rounds of replication.This means that Fragment I must NOT contain a centromere, and so choice A and choice C are incorrect. In Experiment 3B,however, the yeast with Fragment II DO produce visible colonies after 20 rounds of cell division. Therefore Fragment II mustcontain a centromere, and so choice B is correct. Finally, choice D is incorrect because if the mutant yeast’s chromosomes didnot contain centromeres, the mutant yeast strain could not live under any circumstances. Chromosomes without centromereswould be lost during cell division, and all of the genetic information contained on the chromosomes would likewise be lost.And without the genetic information, life is not possible.



MCAT

12 as developed by

8. The correct answer is choice D. This question uses the Roman numeral format and requires some information fromthe passage. The passage states that both microtubule assembly and disassembly are necessary for proper chromosome andplasmid segregation. This is exemplified by the chromosome movements that occur in metaphase of mitosis. Duringmetaphase, the chromosomes, which are attached to microtubules at the centromere, move back and forth across the midline

of the cell. The microtubules are capable of moving the chromosomes in a back and forth fashion because the microtubulesfluctuate between growing longer and then shorter. Because the passage tells you that chromosome segregation requiresmicrotubule assembly and disassembly, you can conclude that the experimental conditions described in both Roman numeralsI and II would disrupt proper chromosome segregation. If chromosome segregation is disrupted, the cells cannot successfullydivide and grow, and if these cells were grown on minimal medium, the result would be the same as for Experiment 1--noincrease in cell number after the first division, and no visible colonies after 20 divisions. The cells in Experiment 1 cannotdivide and grow on minimal medium because they have no source of adenine for DNA synthesis. Cells grown using theexperimental conditions described in Roman numerals I and II would not grow because the microtubules in these cells couldnot properly perform their roles in chromosome segregation. So, because we know that both Roman numerals I and II arecorrect, we can eliminate choices A and B. Let’s look at Roman numeral III. Given that cytokinesis is the final stage of celldivision, adding a cytokinesis inhibitor to the growth medium would also prevent cell growth on minimal medium, or anykind of media for that matter, and would give the same results as seen for Experiment 1.

9. The correct answer is choice C. This question requires that you combine information contained in the passage with

outside knowledge. The passage states that the mutant yeast cells cannot grow unless adenine is supplied in their growthmedium, which implies that this strain has a defect in the adenine biosynthesis pathway. The cells used in Experiment 1 didnot have adenine+ gene-containing plasmids inserted in them, and as we see in Experiment 1, these cells cannot grow onminimal medium. So from the information contained in the passage we can conclude that the mutant cells cannot synthesizeadenine and that is why the mutant cells are unable to grow on minimal medium.

Now that we know this, all we have to do is figure out what role adenine plays in cell life. Adenine is a nitrogenbase contained in DNA and RNA. You should also remember that DNA and RNA are nucleic acids. Thus, the mutant cellsused in Experiment 1 cannot grow because they do not have a source of adenine and thus they cannot synthesize the DNA andRNA that is needed for cell division to occur. Therefore choice C is the correct answer. Let’s look at the wrong answerchoices quickly. Choice A is incorrect because adenine is not a protein or protein precursor. Amino acids are the buildingblocks of proteins, not nucleotides. Choice B is incorrect as well. Fatty acids are the building blocks of certain lipids such asfats or triacylglycerides. Nothing in the passage suggests that the mutant yeast cells have an inability to synthesize fattyacids. Finally, choice D is incorrect because, as is specified in the question stem, minimal medium contains salts.Furthermore, living organisms do not synthesize salts.

10. The correct answer is choice A. Let’s start analyzing this question by predicting what the growth curve should be forthe cells used in Experiment 3B. The cells in Experiment 3B will act like normal, wild-type, yeast cells cultured in growthmedium. You know this because as evidenced in Table 1, these cells continue to produce visible colonies at their 20th celldivision. Furthermore, you should recall that the growth curves of microorganisms are sigmoidal, or S-shaped, like the curvesshown for Experiment 3B in choices A and C. You should have immediately eliminated choices B and D because forcontinuously exponential growth curves like those depicted in B and D to occur, the yeast would have to continuously growexponentially, never running out of nutrients, and never having to prepare metabolically for growth. Obviously, this is not inthe realm of the possible. Let’s briefly review why the growth curves for microorganisms are S-shaped. When cells are firstinoculated onto growth medium and the cell number is very low, the cells must activate the biochemical machinery requiredfor growth and cell division. While the cells are gearing up for growth and cell division, the culture is said to be in the lagphase of growth. Growth during this period is occurring at a very low rate. Once the cells have completed the lag phase, thecells begin to increase their numbers exponentially, which results in the log phase of growth (log is short for logarithmic).This period of exponential growth is a consequence of the fact that each cell in the population divides into two cells. At acertain point, the cells will enter the stationary phase of growth. During this phase, the growing microorganisms havedepleted the growth medium of an essential nutrient, or toxic byproducts of metabolism begin to accumulate, and the growthrate becomes very low, as in the lag phase.

Anyway, back to the question. So, we know that the correct answer is either choice A or C. How do these twographs differ from each other? In choice A, the cells in Experiments 2 and 3A reach a higher cell number after 20 celldivisions than the cells in Experiment 1. In choice C, the cells in Experiments 1, 2, and 3A all have a similar small cellnumber after 20 cell divisions. Well, if you look at Table 1 you will realize that choice A is correct. The cells in Experiments2 and 3A will reach a higher cell number than the cells in Experiment 1, because the cells in Experiments 2 and 3A have anadenine+ plasmid. Even though the adenine+ plasmid has no centromere, it will most likely be maintained through a small




KAPLAN 13

number of cell divisions before being lost and will thus allow the cultures in Experiments 2 and 3A to reach an elevated cellnumber with respect to the culture derived from Experiment 1. As we discussed previously, the mutant cells used inExperiment 1 do not contain an adenine+ plasmid and thus they cannot grow at all on minimal medium. What do I mean by“lost?” Well, according to the passage, plasmids are self-replicating; in other words, they replicate independently of the cellcycle. If the plasmid does NOT have a centromere, during mitosis, it will not be able to bind to the mitotic spindle. Thismeans that when the cell undergoes cytokinesis, there is no guarantee that either of the daughter cells will inherit the plasmidand its copy or copies. In fact, research has shown that such plasmids are inherited only 8-20% of the time. Anyway, choiceC is incorrect.

11. The correct answer is choice D. To answer this question you must understand what is occurring in Experiment 2. InExperiment 2, the mutant yeast cells contain an adenine+ plasmid that does not contain a centromere. Since the plasmid doesnot contain a centromere, the replicated plasmid will not be able to bind to the microtubules of the mitotic spindle, and thusthe replicated plasmid will randomly segregate to daughter cells. By random segregation we mean that since the plasmidCANNOT segregate to daughter cells by normal means, when the cells divide, the plasmid will get distributed to daughtercells by chance, or not at all, and will not be faithfully distributed to ALL daughter cells. Due to this random segregation, theplasmid will eventually be lost from almost all progeny cells. However, after one round of cell division, most of the yeastcells will still have plasmids, and these cells will be able to complete a second round of cell division, thereby furtherincreasing the cell number of the culture. This explains the increase in cell number after one round of division observed in thecells of Experiment 2. [For more detail on how a plasmid gets lost, refer to the explanation to Question 10.]

On the molecular level, random segregation of a plasmid means that the replicated plasmid segregates to daughtercells without binding to the mitotic spindle. As stated in the passage, a plasmid must contain a centromere in order to bind tomicrotubules. Thus choices A and B are incorrect. We can also immediately rule out choice B because the cell divisiondescribed in the passage is mitotic, not meiotic. Meiotic cell division would require mating and sporulation, and the passagegives us no reason to conclude that such processes are occurring. Finally choice C is also incorrect. Microfilaments arecomposed of actin molecules and are involved in formation of the contractile ring during cytokinesis, but are not involveddirectly in the segregation of genetic material via binding to chromosomes or plasmids.

Discretes (Questions 12-16)

12. The correct answer is choice D. From the question stem you know that the transcription factor will bind to nucleicacids. This means that SP1 will bind to both DNA and RNA. So all you have to do to answer this question correctly is figureout which organelle does not contain DNA or RNA. Well you should know that the nucleolus is the small dense region of

the nucleus where rRNA synthesis occurs. Thus, SP1 would bind to material within the nucleolus and so choice A must beincorrect. The mitochondria is the organelle, bound by a double membrane, in which the reactions of the Krebs cycle, electrontransport, and oxidative phosphorylation take place. Mitochondria are the site of the production of the majority of the ATPmolecules produced during aerobic respiration. In addition to this function, mitochondria contain their own DNA, whichmeans that SP1 will bind to it. Therefore choice B is also incorrect. Choice C is also incorrect. Ribosomes are smallorganelles composed of protein and rRNA and are the site of translation. Therefore SP1 would also bind to ribosomes. Sochoice D is our correct answer. The Golgi apparatus is the organelle responsible for the processing, packaging, anddistribution of proteins. It is not composed of nucleic acids, nor does it process them. Since there are no nucleic acids withinthe Golgi apparatus, SP1 would not bind to it. Therefore, choice D is the correct answer.

13. The correct answer is choice C. In order to answer this question correctly you needed to know the functions of eachof the subcellular structures listed as choices. Let’s go through each one until we find one that is NOT responsible fordirecting proteins to their final destinations within cells. The Golgi apparatus and endoplasmic reticulum sort proteinsaccording to their destination using signals inherent in the amino acid sequence of proteins. For example, the amino acid

sequence lys-asp-glu-leu at the N-terminus causes a protein to be transported from the Golgi apparatus to the endoplasmicreticulum. So it is obvious that both the endoplasmic reticulum and the amino acid sequence of a protein play a role inprotein transport. Therefore choices A and D are incorrect. Vesicles, choice B, are formed from the fusing of membranesurfaces. They are involved in transporting proteins from one location to another by virtue of their ability to move around thecell and through membranes. They may fuse with a target membrane and release the protein at its destination. So choice B isalso incorrect. Therefore choice C must be the correct answer. Lysosomes are rather small, spherical membrane-enclosedbodies that contain hydrolytic enzymes. If a protein were to come in contact with these enzymes it would be degraded. Solysosomes are not involved in protein trafficking. Therefore, choice C is the correct answer.



MCAT

14 as developed by

14. Choice A is the correct answer. The key information in the question stem is that this protein, HAP2, enters thenucleus and a molecule to ATP is consumed in the process. All of the other information is just a distraction and I hope that itdid not throw you off. Furthermore, the specific name of the protein--HAP2--is not important in answering the question, andyou are not expected to know anything about this molecule. So all you have to do to answer this question is figure out whichtype of transport uses ATP. Well, passive diffusion is the net movement of dissolved particles down their concentration

gradient from a region of higher concentration to a region of lower concentration. This process does not require ATP, sochoice C is incorrect. Facilitated diffusion is the net movement of dissolved particles down their concentration gradient withthe help of carrier molecules. This process, like passive diffusion, does not require ATP. Therefore, choice B is also incorrect.Active transport is the net movement of particles AGAINST their concentration gradient with the help of carrier molecules.Unlike facilitated diffusion or passive diffusion, active transport requires energy. And as you know, ATP stores energy so thatcells can perform various functions. Thus, choice A is the correct answer. Endocytosis, choice D, is a process in which thecell membrane invaginates, forming a vesicle that contains extracellular medium. So this process, though ATP-dependent,would not account for a cellular protein moving from the cytoplasm into the nucleus. Thus choice D is incorrect.

15. The correct answer is choice C. To answer this question you need to determine the concentration of the solution inthe beaker. How can you do this? Well if you compare the solution levels in the three funnels in the initial state to thesolution levels in the final state, you can determine the concentration of urea in the beaker. Since you know from the questionstem that the funnels are covered with a semi-permeable membrane that is impermeable to urea, the urea itself will not beable to flow from the solutions in the funnels to the solution in the beaker or vice versa. This means that water is the only

substance that will move between the funnels and the beaker; in other words, osmosis will occur. Osmosis is the movementof water across a semi-permeable membrane from a region of lower solute concentration to a region of higher soluteconcentration, until the solute concentrations on both sides of the membrane are equal.

So now that we know that we’re talking about osmosis, let’s look at the diagram. The distilled water funnel had adecrease in water. For the water to move from the funnel to the solution, the solution must have a higher solute concentrationthan distilled water. Therefore, choice D is incorrect. The level of solution in the 1 M urea solution remains unchanged; therewas no net movement of water. This means that the urea concentration in the beaker must be equal to the urea concentrationin the funnel. In other words, the solution in the beaker is isotonic to the solution in the funnel. Therefore, choice C must bethe correct answer. The level of solution in the 2 M funnel increases, indicating that water has moved from the beaker into thefunnel. Thus, the concentration of urea must be greater in the funnel, and so the concentration of urea in the beaker must beless than 2 M. Thus, choices A and B are incorrect.

16. Choice A is the correct answer. This question is about cellular adhesion and recognition. This refers to the adhesionof cells to one another and to the recognition of molecules that interact with the cell, such as hormones, antibodies, and

viruses. So from this you know that we’re dealing with phenomenon of the cell surface, not the cell interior. With this inmind you can eliminate choices C and D, since they do not refer to elements of the plasma membrane related to its exterior.Choice C refers to the side of the plasma membrane that is in contact with neither the exterior cell surface nor the cytoplasm.The fluidity of the plasma membrane, choice B, refers to the fact that both lipids and proteins are able to diffuse throughoutthe entire plasma membrane. This fluidity enables the lipid bilayer to act as a solvent for membrane proteins. Although somemembrane proteins do play a role in adhesion and recognition, MOST do not, so they would only play a minor role in theactual adhesion and recognition process. Thus, choice B is incorrect. So by the process of elimination, choice A must be thecorrect answer. Let’s see why. In order for a cell to adhere to another cell, or for many molecules to interact with a cell, theymust first bind to receptors or ligands on the cell surface. Therefore, carbohydrate molecules on the exterior of the plasmamembrane are in a perfect position to interact with other cell and molecules, and in fact they do. Thus, choice A is the correctanswer.

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