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MCAT > Chemistry > Flashcards

Chemistry Flashcards

1
Q

peptide bond

A

peptide bond = amide bond

peptide bond is formed by the amine group attacking the carbonyl carbon

water is lost during peptide bond formation (hydrolysis)

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2
Q

Metric System Units Mnemonic

A 
King                    Kilo 10^3
Henry                  Hecto 10^2
Died                    Deca 10^1
Unexpectedly     -UNIT-
Drinking              Deci 10^-1
Chocolate           Centi 10^-2
Milk                     Milli 10^-3
Monday              Micro 10^-6
Night                   Nano 10^-9
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3
Q

sp2-hybridized carbon atoms

A

AX3 systems that possess exactly one doubly bonded atom and are therefore sp2-hybridized

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4
Q

Michaelism Mentin

A
  • describes relationship b/t reaction rate (Vo) and substrate concentration [S]. also allows us to estimate the affinity, Km, and Vmax
  • assumes that the catalytic step is the rate-limiting step AND the [ES] rapidly reaches the steady state=stays constant

V = (Vmax [S)] / (KM + [S])

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5
Q

Catalytic constant (kcat)

A
  • How fast ES complex proceeds to E + P
  • Number of catalytic cycles that each active site undergoes per unit time
  • Rate constant of the reaction when enzyme is saturated with substrate

kcat = Vmax/[E]

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6
Q

Absorption of ultraviolet light by organic molecules always results in what process?

A
  • UV light absorption always results in excitation of bound electrons
  • Bond breaking, bond vibration, ionization can occur, but not always
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7
Q

Column chromatography

A

Silica gel is polar and stationary in the column
The mobile phase (e.g. benzene) can be adjusted for polarity to maximize separation.

  • The least polar/most nonpolar compound will always elute first
  • hydrocarbons are very nonpolar & elute first then ketones, alcohols, and carboxylic acids last
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8
Q

The half-life of a radioactive material is

A

the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei (which may or may not be radioactive)

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9
Q

The half-life of a radioactive material is

A

the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei (which may or may not be radioactive)

  • The more unstable something is, the shorter the half-life
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10
Q

Alpha decay

A

Alpha decay: (238/92)U → (4/2)He + (234/90)Th

  • Ejection of a helium nucleus at relatively low speed
  • Weakest form of radiaton
  • Alpha particle: (4/2)a
  • Alter the state of the original element in a way that effectively categorizes it as another element
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11
Q

Beta decay

A

Beta decay: (234/90)Th → (0/-1)e) + (234/91)Pa

  • Ejection of a high energy electron from the nucleus
  • More energy than alpha particles
  • Beta particle: (0/-1)B
  • Alter the state of the original element in a way that effectively categorizes it as another element
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12
Q

Gamma emission

A

Gamma emission: (222/86)Rn → (222/86)Rn + (0/0)Y

  • Ejection of a high energy electron from the nucleus
  • Most energy
  • Gamma ray: (0/0)Y
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13
Q

Half-life vs. [Stuff]

A
  • After 1 half-life, the amount of the original stuff decreases by half.
  • After 2 half-lives, the amount of the original stuff decreases by a factor of 4.
  • After 3 half-lives, the amount of the original stuff decreases by a factor of 8.
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14
Q

Why must a person sitting straight up either lean forward or slide their feet under the chair in order to stand up?

A

as the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person’s center of mass) and the distance along the horizontal between the center of mass and the support point

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15
Q

How where do kinases act on ATP?

A

three phosphate groups, in order of closest to furthest from the ribose sugar, are labeled alpha, beta, and gamma

Kinases transfer the Gamma phosphate to other molecules

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16
Q

Protein stability and melting point

A

higher melting point = more stable protein bc more energy is needed to unfold the protein

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17
Q

Reduction of C=O

A

C-OH

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18
Q

Venturi Effect

:

A

Oxygen pressure= oxygen static pressure (P) + oxygen flow pressure (ρv^2)/2.

  • In the area of the mask openings, P air = P + (ρv^2)/2, thus Pair > P
  • Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask.
  • Venturi tube can be used to determine the velocity of a fluid that is flowing within it
  • Pitot tube, which is used to determine the velocity of a fluid flowing past it.

The continuity equation (Q = Av) states that for a fluid with constant flow rate Q, a decrease in cross-sectional area A is associated with an increase in velocity v.

-Bernoulli’s equation states that for a fluid at a constant height, an increase in velocity is associated with a decrease in pressure.

The decrease in pressure occurs when a fluid flows into a constricted region of a pipe is known as the Venturi effect. A venturi meter can be sued to determine the velocity of a fluid in a pipe.

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19
Q

Bernoulli’s equation

A

The equation resulting from applying conservation of energy to an incompressible frictionless fluid:
P + ½ρv2 + ρgh = constant

  • fast fluid flows have low static pressure
  • High speed flow = high dynamic pressure
  • For a fluid at a constant height, an increase in velocity is associated with a decrease in pressure
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20
Q

DNA bonding and melting point

A

G-C base pairs have triple bonds- 1 sigma and 2 pi bonds
A-T base pairs have double bonds- 1 sigma and 1 pi bond

more pi stacking interactions = most stable thermal stability = high melting point of DNA rich in GC content

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21
Q

relative vapor pressure

A

when the atmospheric pressure=vapour pressure of the substance that is when it will start to escape into the gas phase. Since the atmospheric pressure is lower, that also means that the boiling point is lower. Vapour pressure and boiling point are directly related.

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22
Q

SI units

A

meter, kg, second, ampere, mole, kelvin, candela

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23
Q

Volumetric flow rate (continuity equation)

A

(Q = Av) states that for a fluid with constant flow rate Q, a decrease in cross-sectional area A is associated with an increase in velocity v

OR

Q=V/t, where Q is flow rate, V is volume, and t is elapsed time

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24
Q

continuity equation

A
  • works under the assumption that the flow in will equal the flow out
  • applies to any incompressible fluid

A1∗v1=A2∗v2 where v is the velocity of the fluid and A is the area of the cross section of the space the fluid is moving through

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25
Q

triacylglycerols are

A

neutral storage lipids that consist of three fatty acids ester-linked to a single glycerol

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26
Q

Cathode and Anode Mnemonic

A

PANiC
Positive is Anode
Negative is Cathode

An Ox= Anode Oxidation
Red Cat= REDuction CAThode
-anode shoots out e-
cathode takes in e-

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27
Q

Calculating the Gibbs Free Energy from EMF

A

ΔG° = −nFE°

where n is the moles of electrons transferred, E° is the standard state EMF, F is Faraday’s constant (10^5 C/mol x e-)

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28
Q

Calculate log easy

A

-log(1x10^-5) = 5

10^-5 = 1x10^-5

  • log 2x10^-X = [X-1].7
  • log 3x10^-X = [X-1].5
  • log 5x10^-X = [X-1].3
  • log 8x10^-X = [X-1].1
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29
Q

Current (Faraday’s Law)

A

Current = coulombs of charge per second. I = q/t

Faraday’s constant = coulombs of charge per mol of e- = total charge over total mols of e-. F = q/n.
q = It and q = nF, thus we get:
It = nF
Current x time = mols of e- x Faraday’s constant.

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30
Q

Boyle’s Law

A

Boyle’s law: P1/V1=P2/V2
- constant temperature

**PV=nRT can be divided by T by both sides, leaving PV/T=nR ( a constant )

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31
Q

Charles’s Law

A

Charles’s Law states that V1/T1=V2/T2

- constant pressure

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32
Q

Heisenber’s Principle

A

states as we know more about momentum of a particle the less you know about its position and vice versa

it is impossible to determine both the position and the velocity of an object simultaneously.

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33
Q

Gay-Lussac’s Law

A

P1/T1=P2/T2

- volume is constant

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34
Q

Combined gas law

A

combined gas law:

(P1V1)/T1=(P2V2)/T2

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35
Q

Lewis base

A

a substance that donates an electron pair in forming a covalent interaction

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36
Q

Lewis acid

A

accepts a pair of electrons from a Lewis base (Lewis bases donate electrons)

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37
Q

Broasted-Lowry acid

A

donates a proton to a Bronsted base (Bronsted bases accept protons)

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38
Q

Brownsted-Lowry base

A

Bronsted bases accept protons

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39
Q

Arrhenius acid

A

increases H+ concentration in water (releases an H+)

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40
Q

acid-base reaction

A

H+ is being produced in the forward reaction, and consumed in the reserve reaction

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41
Q

Strong acids

A

HCl, HBr, HI, HNO3, H2SO4, HClO4

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42
Q

Strong bases

A

LiOH, NaOH, KOH,Ca(OH)2, Sr(OH)2, Ba(OH)2,

43
Q

catalysts

A

increase the rate of chemical reaction, which is the amount of product formed per unit time

44
Q

K > 1

A

ΔG° is negative and the reaction is spontaneous

45
Q

K > 1

A

ΔG° is positive and the reaction is nonspontaneous

46
Q

Coordinate covalent bond

A

Lewis acid–base interaction between a metal cation and an electron pair donor

47
Q

Adding HCl to an acid-base rxn

A

NH3 is a stronger base than H2O, so HCl will most likely protonate NH3 creating NH4, thus decreasing [NH3]

48
Q

Le Châtelier’s principle

A

Add reactant=shift right
Add product= shift left
Remove reactant= shift left
Remove product= shift right

With pressure, the reaction is shifted to the side with fewer moles of gas

49
Q
  • add reactants, we want to shift to the R to use up that extra product= equilibrium
  • remove products, we shift to the right to increase production of products

if we add products, we want to shift to the R to decrease [reactant] = equilibrium

remove reactants, we shift to the right to increase production of reactants

A

ADD a product = shift reaction AWAY

Remove a reactant = increase reactant production = shift towards lost side

50
Q

Keq vs. Q

A

Keq utilizes the concentrations of species at equilibrium, Q tells us the reaction’s current position. Thus, if Q is equal to Keq, then the reaction is at equilibrium. However, if Q is less than Keq, it has not yet achieved equilibrium, and the reaction will proceed forward (towards the products). If Q is greater than Keq, then the reaction has gone past equilibrium and will proceed in the reverse direction (towards the reactants).

51
Q

Which of the following atoms will be expected to have the smallest second ionization energy?

A.Na
B.C
C.O
D.Ca

A

D. Ca
Metals have lower ionization energies than non-metals as long as the ionization event involves a valence electron.
- Na is an alkali metal=has only one valence e- and has a large second ionization energy. Ca is an alkaline earth metal and has two valence electrons. It will therefore have the smallest second ionization energy of the four atoms listed, which include Na and two non-metals.

It’s asking about the smallest SECOND ionization energy. Remember that atoms like having a complete octet. Calcium has two valence electrons, so when it is ionized for the first time, it only has one more valence electron between it and a complete octet. If you think of it like that, it wouldn’t take a lot of energy for Ca to give up that last electron to reach noble gas configuration, so it has the smallest second ionization energy.
Edit:: after Na ionizes once, it’s essentially a noble gas and it would take A LOT for it to lose an electron

Ionization energy is the amount of energy required to remove an electron from an atom or an ion. Ionization energy increases as you go up and to the right of the periodic table. When you remove the first electron from an atom, the amount of energy required is called the first ionization energy.

If you want to remove a second electron, the amount of energy required to remove the second electron is called the second ionization energy.

52
Q

mass percent

A

(coefficientt of atom in the formula of the substance x molar mass) / molar mass of substance

ex) oxygen in water- (1 x 16) / 18 = 89%

53
Q

Henderson-Hasselbach equation

A

calculate the pH of the solution

pH = pKa + log([base]/[acid])

54
Q

structure of the DNA double helix

A

Sugar-phosphate backbones form the exterior of the double helix
Interior helix:
- purines bind w/ pyrimidines
- Dan stands are antiparallel (opposite directions)
- base pairs (G-C) appear in equal concentrations

55
Q

calculate the number of charges in excess

A

formula for charge is q = ne
q is the amount of charge (in Coulombs)
n is the number of electrons
e is the elementary charge (1.6 × 10^–19 C)

56
Q

power consumed

A

P=PE /time = mgh/t

PE= mgh = mass x gravitational constant x distance from earth’s surface

57
Q

Gativational potential energy and Work done against gravity

A

Wgrav = F x d = m x g x Δh = ΔPEgrav

58
Q

Ways to calculate work

A

Work done by a constant force= F x d x cosθ = -pΔV = ΔE

Work (kinetic energy) Wnet= ΔKE
- net work on a system = change in KE of moving things

Work done against gravity Wgrav = F x d = m x g x Δh

KE=1/2mv^2

P=W/t = ΔE/t

If pressure is constant, W = P/ΔV (P= length of rectangle under the Pressure volume curve and width is ΔV)

(Springs and rubber bands) Elastic PE= U = 1/2kx^2 (k is the spring constant and x is the magnitude of displacement)

59
Q

Ways to calculate work

A

Work done by a constant force= F x d x cosθ = -pΔV = ΔE

Work (kinetic energy) Wnet= ΔKE
- net work on a system = change in KE of moving things

Work done against gravity Wgrav = F x d = m x g x Δh

KE=1/2mv^2

P=W/t = ΔE/t

If pressure is constant, W = P/ΔV (P= length of rectangle under the Pressure volume curve and width is ΔV)

(Springs and rubber bands) Elastic PE= U = 1/2kx^2 (k is the spring constant and x is the magnitude of displacement)

60
Q

Intensity of sound

A

β = 10 logI/I0

  • β is sound level in decibels, I is intensity. I0 is 10-12 W/m2
  • Intensity is power per area, or the rate of energy expenditure per area. The unit is W/m2
61
Q

Intensity of sound ratio

A

β = 10 logI/I0

  • β is sound level in decibels, I is intensity. I0 is 10-12 W/m2
  • Intensity is power per area, or the rate of energy expenditure per area. The unit is W/m2
Ratio of sound intensities: 
B2 = B1 + 10log (I2/I1)
(B2-B1) 10 = # = log I2/I1
take 10 to the power on both sides to cancel log
10^# = I2/I1
62
Q

Lenses and glasses

A

Reading glasses (“farsighted”) = converging lenses

normal glasses (“nearsighted”) = diverging lenses

63
Q

Half life equation

A

amount of original material = amount of original material left x 2 ^#half-lives = amount of original material left x 2 ^1/half life

64
Q

protein secondary structure

A

represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms.

65
Q

resistors in parallel

A

1/Rtot=1/R1+1/R2+…

Req for 2 resistors in parallel is = R1*R2/(R1+R2)

66
Q

the primary reason for chemical separation during a chromatographic separation

A

the relative amount of hydrogen bonding to the stationary phase will determine the relative rate of migration of the various components in the sample

67
Q

Rf

A

the ratio of the distance travelled by the analyte relative to the solvent front during a chromatographic separation

68
Q

stereospecific

A

provides essentially one stereoisomer product

- SN2

69
Q

stereoselective

A

provides some sort of mixture of chiral products

- SN1

70
Q

SN1

A
  • bimolecular (2 steps)
  • polar protic solvents (can H bond)
  • favors 3>2>1>methyl
  • rate = k[RL]
  • racemic (2 inverse products
  • strong nucleophile not required
  • forms a carbocation (favors cation stability)
  • barrier = carbocation stability
71
Q

SN2

A
  • unimolecular (1 step)
  • polar aprotic solvents (can’t H bond)
  • favors methyl > 1 > 2> 3 C
  • rate = k[Nu][RL]
  • inverted products; optically active
  • strong nucleophile favored
  • barrier = steric hindrance
72
Q

amide group

A

CO=NH2

73
Q

amine

A

R-NH2

74
Q

aldehyde

A
  • CHO

- C=O and C-H

75
Q

ketone

A

-COC-

C=O with r groups on both side

76
Q

ester

A
  • COOR’
  • no H holding
  • insoluble in water
77
Q

imine

A

C=N

78
Q

enamine

A

N-C=C

79
Q

product formed for aldol condensation

A

β-hydroxy aldehyde or β-hydroxy ketone.

80
Q

aldol addition product

A

hydroxy ketone or aldehyde

81
Q

anhydride

A

-O=C-O-C=O-

82
Q

first peak in the gas chromatograph (gc) trace

A

lowest molecular weight = weakest intermolecular forces of attraction = migrates the fastest = first peak

83
Q

carboxylic acid derivatives as leaving groups

A
  • the carbonyl oxygen is the one that is double bonded to the carbon and is not likely to be a good leaving group in this reaction
  • oxygen on the -OH group is more likely to leave
84
Q

focal length & strength

A

1/f = 1/o + 1/i = 2/r

Strength (Power) = 1/f

= S = 1/o + 1/i

  • meters
    O (the object) = 1m (this is the distance between the object and the lens of your eye) We are also given I (the image) is 2cm away on the other side of the lens (your retina - which is where the image forms). 2 cm =

2cm*1m/100 cm = 1/50 m = 0.02m

85
Q

Ohm’s law

A

V=IR = I= V/R
current is equal to voltage divided by resistance. If current increases and resistance is constant, then voltage increases as well.

86
Q

index of refraction of a medium

A

index of refraction of a medium is equal to the ratio of the speed of light in vacuum to the speed of light in the medium

87
Q

intensity of electromagnetic radiation

A

the energy of electromagnetic radiation is directly proportional to the number of photons

the intensity of electromagnetic radiation is defined as energy emitted per unit time. Thus, intensity is directly proportional to the number of photons emitted.

88
Q

THz to Hz

A

1 THz = 10^12 Hz

89
Q

flow rate

A

mL/min

90
Q

binding regions/amino acids

A
  • hydrophobic region is that CH2 region
  • Cross linking residues are cysteines
  • Spacers would be the glycines
  • You can tell the phosphorylated residue by the way that it is.
  • Arg and Asp which facilitate cell adhesion.
91
Q

Why is the velocity of blood flow slower in capillaries than in arteries?

A

the high number of capillaries in the body means that the total cross-sectional area of these vessels is larger than any other vessel type in the circulatory system. This causes the velocity of the blood to decrease

92
Q

hybridization of an octahedral compound

A
  • octahedral compounds have six σ bonds and no lone pairs. - the central atom requires the hybridization of six atomic orbitals, d2sp3
93
Q

assumptions of the ideal gas law

A

a property of an “ideal” gas is that it is composed of particles that have negligible volume and do not exert intermolecular forces.

94
Q

molecules made mostly of C and H

A

nonpolar- hydrophobic

95
Q

the requirement of ATP for kinase activity implies . . .

A

the phosphate groups come from ATP

96
Q

AX3 systems

A
  • have 3 bonded atoms and no lone pairs
  • preferred geometry for such a system is trigonal planar and the hybridization scheme that facilitates this geometry is sp2.
97
Q

lipase

A

hydrolyze fatty acids

98
Q

glycogen synthase

A

the main chain linkage of glycogen, which is an α-1,4-glycosidic bond

99
Q

removal of phosphate prior to gel electrophoresis

A

result in a smaller band corresponding

100
Q

atomic number and mass number

A
  • number of electrons in a neutral atom is equal to the number of protons
  • mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus
  • the number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z) (Number of electrons)
101
Q

Which is less likely to dissociate in water?

HPO42−, H2SO4, H3PO4, HCL

A

HPO42− has a high negative charge and so dissociation of it will occur to the least extent

102
Q

focal length

A

According to the thin lens equation, the distance between the lens and the image is 3F x 1F/(3F - 1F) = (3/2)F, where F is the focal length. The ratio sought is then equal to (3/2)F/(3F) = 1/2.

103
Q

E° and spontaneity

A

spontaneous reaction = E° > 0

non spontaneous reaction = E° < 0

104
Q

transmission of mutations

A

only occurs in germ cell lines, NOT somatic

Ex) a mutation only expressed in tumor cells won’t be passed down to kids

MCAT (4 decks)

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