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AAMC MCAT Practice Exam 1 With Correct Answers.

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AAMC MCAT Practice Exam 1 With Correct Answers.

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  • August 8, 2024
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Lectphilip
AAMC MCAT Practice Exam 1 With
Correct Answers

In nnthe nnchromatography nnof nnthe nnreaction nnmixture, nnwater nnabsorbed nnon nncellulose nnfunctioned nnas
nnthe nnstationary nnphase. nnWhat nnwas nnthe nnprincipal nnfactor nndetermining nnthe nnmigration nnof nnindividual

nncomponents nnin nnthe nnsample?




A) nnHydrogen nnbonding

B) nnSolute nnconcentration

C) nnStationary nnphase nnconcentration

D) nnThickness nnof nnpaper nn- nnANSWER: nn✔ nnA) nnHydrogen nnbonding nn



The nnanswer nnto nnthis nnquestion nnis nnA nnbecause nnthe nnrelative nnamount nnof nnhydrogen nnbonding nnto nnthe
nnstationary nnphase nnwill nndetermine nnthe nnrelative nnrate nnof nnmigration nnof nnthe nnvarious nncomponents nnin

nnthe nnsample.




What nnassumption nnis nnbeing nnmade nnif nnscientists nnconclude nnthat nnaspartic nnacid nnwas nnformed nnby nnthe
nnprebiological nnsynthesis nnin nnthe nnpassage?




A) nnAspartic nnacid nnis nnunstable nnat nntemperatures nnbelow nn150 nndegrees nnC. nn

B) nnAll nnof nnthe nnmalic nnacid nnunderwent nnthe nndehydration nnreaction nnto nnform nnfumaric/maleic nnacid. nn

C) nnCompound nnA nnand nncyanide nnwere nnavailable nnon nnprimitive nnEarth. nn

D) nnThe nnreaction nnbetween nnammonia nnand nnfumaric nnacid nnwas nncatalyzed nnby nnthe nnpresence nnof nnwater.
nn- nnANSWER: nn✔ nnC) nnCompound nnA nnand nncyanide nnwere nnavailable nnon nnprimitive nnEarth. nn




he nnanswer nnto nnthis nnquestion nnis nnC nnsince, nnin nnorder nnfor nnthe nnexperimental nnreaction nnsequence nnto
nnbe nnrelevant nnto nnthe nnprimordial nnformation nnof nnaspartic nnacid, nnthe nnstarting nnmaterials nnused

nn(Compound nnA nnand nncyanide) nnare nnassumed nnto nnhave nnbeen nnavailable.

,According nnto nnthe nndeveloped nnchromatography nnplate nnshown nnbelow, nnwhat nnis nnthe nnapproximate nnRf
nnvalue nnof nnaspartic nnacid?




aspartic nnacid nn= nn2

solvent nnfront nn= nn10 nn



A) nn0.20

B) nn0.50

C) nn5

D) nn10 nn- nnANSWER: nn✔ nnThe nnanswer nnto nnthis nnquestion nnis nnA nn(0.20) nnbecause nnRf nnis nnthe nnratio nnof nnthe
nndistance nntravelled nnby nnthe nnanalyte nnrelative nnto nnthe nnsolvent nnfront nnduring nna nnchromatographic

nnseparation. nnAspartic nnacid nntravelled nntwo nnunits, nnwhile nnthe nnsolvent nnfront nntravelled nnten nnunits,

nngiving nnan nnRf nnof nn2/10 nn= nn0.20 nnfor nnaspartic nnacid.




Which nnof nnthe nnfollowing nnstatements nndoes nnNOT nncorrectly nndescribe nnthe nndehydration nnof nnmalic nnacid
nnto nnfumaric nnacid nnand nnmaleic nnacid? nn




A) nnThe nnreaction nnoccurs nnmost nnreadily nnwith nntertiary nnalcohols. nn

B) nnThe nnreaction nninvolves nnthe nnloss nnof nna nnwater nnmolecule. nn

C) nnThe nnreaction nnhas nna nncarbocation nnintermediate.

D) nnThe nnreaction nnis nnstereospecific. nn- nnANSWER: nn✔ nnD) nnThe nnreaction nnis nnstereospecific.



The nnanswer nnto nnthis nnquestion nnis nnD nnbecause nnthe nnfact nnthat nnboth nnfumaric nnand nnmaleic nnacid nnare
nnproduced nnmeans nnthat nnthe nndehydration nnof nnmalic nnacid nnis nnNOT nnstereospecific.




What nntype nnof nnfunctional nngroup nnis nnformed nnwhen nnaspartic nnacid nnreacts nnwith nnanother nnamino nnacid
nnto nnform nna nnpeptide nnbond? nn




A) nnAn nnamine nngroup nn

B) nnAn nnaldehyde nngroup

C) nnAn nnamide nngroup

,D) nnA nncarboxyl nngroup nn- nnANSWER: nn✔ nnC) nnAn nnamide nngroup



The nnanswer nnto nnthis nnquestion nnis nnC nnbecause nnthe nnfunctional nngroup nnthat nnforms nnduring nnpeptide
nnbond nnformation nnis nnknown nnas nnan nnamide nngroup.




If nn2-pentanol nnreplaces nn1-pentanol nnin nnthe nnreaction nnshown nnin nnFigure nn3 nn(SN2), nnthe nnrate nnof
nnsubstitution nnis nnless nnbecause: nn



A) nnthe nnC-O nnbond nnin nn2-pentanol nnis nnstronger nnthan nnthe nnC-O nnbond nnin nn1-pentanol.

B) nnthere nnis nna nncompeting nnelimination nnreaction nnthat nnslows nnthe nnrate nnof nnsubstitution.

C) nnthere nnis nnmore nnsteric nnhindrance nnat nnthe nnoxygen nnatom nnin nn2-pentanol nnthan nnin nn1-pentanol,
nnmaking nnprotonation nnless nnlikely.



D) nnthere nnis nnmore nnsteric nnhindrance nnat nnthe nn2-position nnof nn2-pentanol nnthan nnat nnthe nn1-position nnof
nn1-pentanol. nn- nnANSWER: nn✔ nnD) nnthere nnis nnmore nnsteric nnhindrance nnat nnthe nn2-position nnof nn2-pentanol

nnthan nnat nnthe nn1-position nnof nn1-pentanol.




The nnanswer nnto nnthis nnquestion nnis nnD nnbecause nnthe nnrate nnof nnsubstitution nnof nnprotonated nnalcohols nnis
nnsubject nnto nnsteric nnhindrance. nnThis nninhibits nnthe nnability nnof nnnucleophiles nnto nncollide nnwith nnthe

nnreacting nnelectrophilic nncenter nnand nnslows nnthe nnrate nnof nnreaction.




If nna nnsolution nncontaining nnthe nncompounds nnshown nnin nnFigure nn4, nnis nninjected nninto nna nngas-liquid
nnchromatograph, nnthe nnfirst nnpeak nnobserved nnin nnthe nngc nntrace nnis nnattributable nnto nnwhich nncompound?



A) nnMethyl-2-butanol

B) nnMethyl-2-butene

C) nnChloro-2-methylbutane

D) nnBromo-2-mehtylbutane nn- nnANSWER: nn✔ nnB) nnMethyl-2-butene



The nnanswer nnto nnthis nnquestion nnis nnB nnbecause nn2-methyl-2-butene nnwill nnexhibit nnthe nnlowest nnmolecular
nnweight nnand nnalso nnthe nnweakest nnintermolecular nnforces nnof nnattraction. nnThis nnsubstance nnwill nntherefore

nnmigrate nnthe nnfastest nnand nnbe nnthe nnfirst nnpeak nnin nnthe nngas nnchromatograph nn(gc) nntrace.




R nnvs. nnS nnstereochemistry* nn- nnANSWER: nn✔ nnR nn= nnclockwise

C nn= nncounter-clockwise

, Acetic nnacid nnand nnethanol nnreact nnto nnform nnan nnester nnproduct nnas nnshown nnbelow. nn



In nndetermining nnwhich nnreactant nnloses nnthe nn-OH nngroup, nnwhich nnof nnthe nnfollowing nnisotopic
nnsubstitutions nnwould nnbe nnmost nnuseful?




A) nnReplace nnthe nnacidic nnH nnof nnacetic nnacid nnwith nnD

B) nnReplace nnthe nnalcoholic nnH nnof nnethanol nnwith nnD. nn

C) nnReplace nnthe nncarbonyl nnoxygen nnof nnacetic nnacid nnwith nnO-18. nn

D) nnReplace nnthe nnhydroxyl nnoxygen nnof nnethanol nnwith nnO-18. nn- nnANSWER: nn✔ nnD) nnReplace nnthe nnhydroxyl
nnoxygen nnof nnethanol nnwith nnO-18.




The nnanswer nnto nnthis nnquestion nnis nnD nnbecause nnthis nnexperiment nninvolves nnlabeling nna nngroup nnwhich
nndoes nnnot nnexchange nnwith nnother nngroups nnpresent nnprior nnto nnreaction nnand nnwill nntherefore nngive

nninformation nnabout nnthe nntrue nnidentity nnof nnthe nngroups, nnwhich nnare nnexchanged nnduring nnthe nnreaction.




A nnperson nnwhose nneye nnhas nna nnlens-to-retina nndistance nnof nn2.0 nncm, nncan nnonly nnclearly nnsee nnobjects
nnthat nnare nncloser nnthan nn1.0 nnm nnaway. nnWhat nnis nnthe nnstrength nnS nnof nnthe nnperson's nneye nnlens? nn(Note:

nnUse nnthe nnthin nnlens nnformula nn1/O nn+ nn1/I nn= nnS) nn




A) nn-50 nnD

B) nn-10 nnD

C) nn51 nnD

D) nn55 nnD nn- nnANSWER: nn✔ nnC) nn51 nnD



nnThe nnanswer nnto nnthis nnquestion nnis nnC nnbecause nnthe nnstrength nnof nnthe nneye nnlens nnis nnequal nnto nnthe
nninverse nnof nnthe nnfocal nnlength nnof nnthe nneye nnlens. nnIts nnnumerical nnvalue nnis nngiven nnby nn(1 nnm)-1+(0.02

nnm)-1=1 nnD+50 nnD=51 nnD.




Which nnstatement nncorrectly nndescribes nnhow nnenzymes nnaffect nnchemical nnreactions? nnStabilization nnof: nn

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