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MCAT > Biology TPR > Flashcards

Biology TPR Flashcards

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MCAT flashcards
1
Q

Bacteria Growth

A

Initially, new culture will undergo LAG PHASE as the bacteria replicate and get ready to divide
Then, binary fission to increase exponentially–LOG PHASE. After some time, it can reach STATIONARY PHASE, where it looks turbid (not clear)

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2
Q

Plaque

A

Clear area on a bacterial plate. Addition of toxin, antibiotic, or virus creates this clear area.

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3
Q

NADH molecules from cellular respiration

A 
1 glucose
Glycolysis: 2 NADH
Pyruvate Dehydrongenase Complex: 2 NADH
Krebs Cycle: 6
Total: 10
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4
Q

Template Strand

A

The strand used for transcription

RNA Polymerase binds

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5
Q

Coding strand

A

Opposite to template strand

NOT used for transcription

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6
Q

F Factor

A

Conjugation between two bacteria–male and female
F factor is the FINAL gene transferred during conjugation. If the F factor doesn’t make it over to the female cell, she will stay female (but acquire new genetic traits). If F factor is transferred, it can end up in a plasmid (becoming F+ MALE), or the genome (cell would be Hfr).

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7
Q

Hfr

A

Hfr strains have a F factor in the genome, and F+ (male) strain has a F factor in the plasmid. BOTH can mate with the female F-

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8
Q

Productive Viral Cycle

A

Virus takes over certain parts of the host cell
Allow production of viral proteins to replicate viral genome
Viruses bud out, and cell doesn’t die
GENERATES VIRUSES WITH OUTER ENVELOPE

(but in Lytic, cell dies)

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9
Q

Attachment and Penetration

A

Can also be called,
Attachment: Adsorption
Penetration: Eclipse

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10
Q

Prokaryotic Viruses vs Animal Viruses

A

Prokaryotic viruses lack an envelope and therefore cannot undergo membrane fusion with host. They rely on injection of viral contents.

Animal viruses can enter their host via membrane fusion or endocytosis. In both cases, receptor binding (specificity) is required, and viral genome is uncoated

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11
Q

Initiation of Translation steps

A

tRNAmet binds to small ribosomal subunit (without mRNA)
mRNA binds to small subunit with the help of 5’ Guanine cap
Small subunit scans mRNA until AUG found
met-tRNAmet associated with mRNA on P site
Large subunit binds

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12
Q

(+) and (-) RNA virus

A

(+) can serve as templates for transcription, but (-), complementary to transcript template, CANNOT be directly translated
Both must encode an RNA-dependent RNA Polymerase to replicate the genome
(-) RNA virus must carry a copy of this protein
(+) is immediately infective

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13
Q

Chemoheterotroph

A

Chemotroph gets energy from Chemicals

Heterotrophs rely on Organic Molecules, not CO2

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14
Q

Auxotroph

A

cannot synthesize particular amino acid

Leucine auxotrophs can’t make leucine

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15
Q

Fermentation Oxidation and Reduction for Lactic acid and alcoholic

A

Regenerates NAD+ from NADH
Lactic acid fermentation: oxidation of NADH occurs via reduction of pyruvate to lactic acid
Alcoholic Fermentation: pyruvate decarboxylated to ethanol

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16
Q

Bacteria vs Eukaryotic flagella

A

Bacteria: Flagellin, uses Rotation powered by proton or sodium gradient

Eukaryote: Microtubules, uses ATP

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17
Q

Transformation

A

Bacteria take in plasmids and DNA fragments and integrates them into genome
When bacteria dies, it lyses and spills DNA into environment. Another bacteria can take them and integrate them

18
Q

Transduction

A

Bacteriophages undergoing lysogenic life cycle incorporate the viral DNA into the bacterial genome

19
Q

Conjugation

A

Bacteria transfer DNA between one another through the sex pilus

Bacteria that can make pilus is F+ (male)

20
Q

Allosteric Regulation

A

Bind allosteric regulator to allosteric site.
Turns it on: allosteric regulator
Phosphorylation (kinases will phosphorylate with high energy phosphate groups)

21
Q

Kinase

A

Regulate enzymes using phosphorylation

22
Q

Km

A

Concentrate of substrate at 1/2 Vmax

Measure of affinity between substrate and enzyme

23
Q

Competitive Inhibition

A

Inhibitor binds to active site
Increase substrate, will reach Vmax eventually. overwhelms inhibitor

decrease affinity (because of inhibitor), Km increase

24
Q

Non-competitive Inhibition

A

Binds at allosteric site.
Even if you add a lot of substrate, Vmax wil be lower.
Same affinity, Km doesn’t change, but enzymes are turned off
Lower Vmax

Always assume reversibility

25
Q

Km increase

A

affinity decrease

26
Q

antisubstrate in enzyme regulation

A

increase Km, but won’t change Vmax

27
Q

Vmax depends on…

A

Total Enzyme concentration

28
Q

Increase enzyme concentration

A

Vmax will change

Km would remain constant

29
Q

Electron Transport/Oxidative Phosphorylation Goals

A
  1. oxidize the electron carriers NADH->NAD+

2. make usable energy

30
Q

PDC Products, and ATP Equivalent

A

2 NADH=5ATP

31
Q

Krebs Cycle Products, and ATP Equivalent

A

6 NADH 15ATP
2 FADH2 3ATP
2 GTP 2 ATP

32
Q

Glycolysis Products, and ATP Equivalent

A

-2ATP
+4 ATP
2NADH 5ATP (2ATP for Eukaryotes)

33
Q

Fermentation

A

No new ATP
Regenerate NAD+ to continue glycolysis
Animals: Pyruvate–>Lactate
NADH->NADH+
Yeast: Pyruvate–>acetyaldehyde–>Ethanol
CO2 produced NAD+

End products are toxic!
Only 2 Net ATP from glycolysis

34
Q

Other foods sources, and processes

A

Amino acids–>PDC, acetyl coA
Fat—Beta-oxidation–>acetylCoA
EtOH—->Acetladehyde–>Krebs
MeOh—->Formaldehyde

35
Q

Transcription factors affecting enzymatic expression is _____ a mechanism of direct enzymatic regulation. It is an _____ method of regulation. Direct enzymatic regulations can be____etc.

A

Transcription factors affecting enzymatic expression is NOT a mechanism of direct enzymatic regulation. It is an INDIRECT method of regulation. Direct enzymatic regulations can be PROTEIN HYDROLYSIS, PHOSPHORYLATION, ALLOSTERIC REGULATION, ASSOCIATION WITH ANOTHER PROTEIN etc.

36
Q

A mutation in the active site of an enzyme would _________

A

A mutation in the active site of an enzyme would affect its Km.

37
Q

Km is the _______ of ________ necessary for a rxn to achieve _______

A

Km is the concentration of substrate necessary for a rxn to achieve 1/2Vmax

38
Q

Due to electron transport in a eukaryotic cell, the pH in the matrix of the mitochondria is _____ because protons are pumped ______ the inner membrane of the mitochondria.

A

Due to electron transport in a eukaryotic cell, the pH in the matrix of the mitochondria is HIGHER because protons are pumped ACROSS the inner membrane of the mitochondria.

39
Q

The _____ the degree of cooperativity of hemoglobin, the more ____ the oxygen binding and release from hemoglobin. With _____ degree of cooperativity, hemoglobin would release oxygen over a ______ range of ______ of oxygen

A

The GREATER the degree of cooperativity of hemoglobin, the more CONCERTED the oxygen binding and release from hemoglobin. With LOWER degree of cooperativity, hemoglobin would release oxygen over a LARGER range of PARTIAL PRESSURE of oxygen

40
Q

Irreversible inhibitors bind ______ to specific enzymes. When bound to the active site, this results in __________

A

Irreversible inhibitors bind COVALENTLY to specific enzymes. When bound to the active site, this results in NEGLIGIBLE PROTEASE ACTIVITY

MCAT (6 decks)

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