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MCAT AAMC Practice Exams/ QP/SB > BB CORRECT > Flashcards

BB CORRECT Flashcards

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MCAT flashcards
1
Q

Phosphatase Function

A

utilizes water to remove molecules of Pi

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2
Q

Michealis Menton

A
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3
Q

L-Isomer are used as a form of proteins during:

A

ribosomal protein synthesis

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4
Q

GTP is a nucleotide: draw it’s structure

A
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5
Q

Bacterial graph

A
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6
Q

Prokaryotes gene expression

A

green

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7
Q

Euchromatin

A

green

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8
Q

if you had 23% A, how much G, C, & T?

A

A & T are equal

G & C are equal

23%=T

27%=G

27%=G

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9
Q

Lac Operon

A

Lac Operon—-> lactose—>transcription

Lac Operon—->No lactose—-> REPRESSION transcription

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10
Q

Lac Operon when glucose is present & absent

A
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11
Q

LacX, LacY, LacZ

A
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12
Q

Cytochrome C

A

GREEN

  • 1 electron carrier, ferrous & ferric state, FADH
  • pumps 4 protons into intermembrane space
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13
Q

Fatty Acid structure

A

you can tell its a fatty acid by the carbonyl group

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14
Q

Bicarb Buffer System:

  • Carbonic anhydrase inhibitor will do what?*
  • PCO2—> inc CO2—->?*
A
  • it will increase CO2 in tissues
  • INC H+ (dec pH)—> inc dissociation of oxygen (dec O2 affinity)
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15
Q

Saturation/Kinetics

A

green

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16
Q

microfilaments are composed of what?

A

ACTIN

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17
Q

thermogenesis happens in what organelle?

A

mitochondria

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18
Q

Lysosomes

Endosomes

A
  • Lysosomes are membrane-bound vesicles that contain acid-hydrolase enzymes (aka hydrolytic enzymes) that function to breakdown various molecules and substances.
  • The acid hydrolases are made in the ER, then modified in the golgi apparatus → golgi apparatus produces vesicles containing these enzymes which will then fuse with an endosome to form a lysosome.

■ Endosomes can also go to the plasma membrane and the trans-golgi network

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19
Q

Fungi is what classification?

A

Eukarya

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20
Q

Non-disjuntion happens during what phase?

A

Meiosis I- Anaphase I

is the failure of homologous chromosomes (M1-anaphase 1) or sister chromatids (M2, anaphase 2)to separate properly during cell division.

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21
Q

Visual pigments absorb lights, what happens if we have fewer pigments?

A

fewer signals of weakly perceived colors will be sent to the brain

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22
Q

a hypothesis/ evidence test

A

you can remove neutrophil to seee if there is a difference

you remove something to see if difference stands still

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23
Q

Photoreceptors

Rods and conse

A

rods: dim-light
cones: color light

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24
Q

Each experiment has:

A

control (baseline) & experimental group (manipulated group)

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25
Q

Veins & capilliaries

A

green

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26
Q

1A: Q 8 : skill 1

AA structure, name A-D

A

a. proline
b. tyrosine
c. histidine
d. phynalanine

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27
Q

1A: Q9: Skill 9

which aa has disuflide linkage?

A

cysteine, cys, C

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28
Q

1A: Q24: skill 1

amino acids

A
  • passage said that HDACs modify basic residues and I choose lysine
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29
Q

1A: Q 27: skill 1

amino acids

A
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30
Q

1A: Q 30: skill 1

allosteric inhibition

A
  • binding of an inhibitor to a site other than the substrate binding site
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31
Q

1A: Q 57: skill 2

feedback inhibition

A
  • feedback inhibition: answer
  • the question notes that the level of glutamine synthetase molecules in the cell remains constant regardless of the tryptophan concentration, thus excluding transcriptional or translational controls. In absence of other types of control, an increase in tryptophan concentration that is accompanied by decreased activity of glutamine synthetase is consistent with feedback inhibition.
  • translational control: if tryptophan inhibited the synthesis of glutamine synthase, but it does not as it is level no matter the W conc
  • transcriptonal control: if tryptophan inhibited the synthesis of glutamine synthase mRNA, but it does not as it is level no matter the W conc
  • By stating that the concentration is constant, the question referring to the ACTIVITY of glutamine synthetase (and NOT the quantity) decreasing as the concentration of tryptophan increases. This excludes transcriptional/translational control because this means that you are not decreasing the number of enzymes present, but the ACTIVITY.

Feedback inhibition is when lots of product is produced and then it prevents further enzyme activity. Item A makes Product B which makes Product C. If the creation of product C somehow limits the activity of product A that would be referred to as negative feedback inhibition (the much more common kind.). If product C’s presence actually increased the activity of Item A then you would refer to it as positive feedback inhibition. ​

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32
Q

1A: Q58 skill 1

competitive inhibition

A
  • competitive inhibition, the inhibitor reversibly binds to the enzyme’s active site
  • all enzymes lover the Ea
  • Noncompetitive inhibitors allosterically bind the enzyme and enzyme-substrate complex with equal affininty
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33
Q

3B: Q1: Skill 2

Muscles

Individual M has longer limbs and muscles than Individual N. Both individuals lift the same amount of weight from the ground to their shoulders an equal number of times. If all repetitions were completed in synchrony, which statement about the amount of work completed by the individuals is true? At the end of the workout:

A
  • Answer: individual N will have done less work because the object was moved a shorter distance
    • because work is equal to the product of the force applied to an object and the distance over which the distance is moved (given in passage) W=Fd (less work, then less force applied and the object is moved at a shorter distance)
    • Individual N has shorter limbs (and likely a shorter ground to shoulder distance), thus this individual will perform less work because the muscles contracted a shorter distance, and the object was moved a shorter distance.

another answer choice was given but it was wrong: Individual M will have done more work because the muscle contracted faster to achieve synchrony with Individual N.​

  • this is a true statment that individual M did more work but work is not dependent on the speed, & thus the rate of muscle contraction is not relevant to the amount of work done by the individuals, it doesn’t matter if the muscle contracted faster
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34
Q

3B: Q2: skill 2: passage based

slow-twitch/fast twitch, same diameter, generating force

A
  • similarily sized fibers would generate similar force, regardless of the fiber type , based on paragraph one becyase force is proportional to cross-sectional area
  • However, since the fast-twitch fiber would generate the same force with greater velocity (contract faster), the fast-twitch fiber would generate more power.
    • ​in addition, the passage mentions that fast twitch fibers “exhibit greater contractile strenght” which leads to an increased rate of ATP hydrolysis, leading to increasing power
  • force is similar: power is disimilar
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35
Q

1 D: Q3: skill 1

metabolizing anaerobically—>fermentation

A
  • fermentation of glucose to lactic acid
    • Remember that glycloysis has a fork in the road at it’s completion. On the one hand it can send Pyruvate to either the citric acid cycle or send it to the fermentation process. If there is no oxygen the fermentation process will recieve the pyruvate
  • depletion of stores of glycogen, this occurs is in order to produce more glucose that can enter lactate fermentation to ultimately make 2 moles of ATP/glucose
  • acidification of the cytoplasm, fermentation is an acidic process that takes place in the cytosol

remember: lack of oxygen effectively shuts down Krebs and ETC so phosphorylation of ADP within the electron transport chain will not occur bc ETC only provides ATP aerobically

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36
Q

3B: Q4: skill4: passage, graph based

work, force, power

A
  • Remeber: W=Fd, you can see that work and force are directly proportional.
    • So if you follow the force curve on the graph, Point A is the lowest force value, and thus the lowest work value.
      • (Also think that since Power = Work/time, Work = Power * time and since time of contraction is the same and you’re given power curve, you can just choose the point with the smallest power value)
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37
Q

1B: Q 7: skill 2

pyrmidines & purines

A
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38
Q

1C: Q 10: skill 2

Evolution: convergent evolution & other types

give examples

A
  • convergent evolution is defined as a process whereby distantly related organisms independently evolve similar traits to adapt to similar needs, example: dolphin & shark
    • dolphins are mammals whereas sharks are not-so they obviously come from different ancestors
    • however, they have similar funtionalities that allow them to survive in the same location
  • divergent evolution: domestc sheep & the mountain goat, polar bear & the panda bear
    • classic example of disruptive selection: light-colored & the dark-colored forms of the peppered moth: (two different phenotypes are maintained because they are better adapted for different environments.) (this is also divergent evolution)
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39
Q

1C: Q11: skill 2: discrete

Assume that a certain species with sex chromosomes R and S exists such that RR individuals develop as males and RS individuals develop as females. Which of the following mechanisms would most likely compensate for the potential imbalance of sex-chromosome gene products between males and females of this species?

A
  • answer: inactivation of one R chromosome in males
    • because males of the species have two copies of the R chromosome, whereas females have one R chromosome and one S chromosome. Therefore, in order to compensate for a potential imbalance of sex chromosomes between females and males, one R chromosome should be inactivated in males. this is anologous to X inactivation of the X chromosome on XX
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40
Q

3A: Q12: skill 2: discrete

GnRH, hormones

A
  • reproductive axis
    • gonadotropin-releasing hormone regulates pituitary gonadotropin (luteinizing hormone and follicle-stimulating hormone) secretion. Thus, if gonadotropin-releasing hormone is not able to regulate luteinizing hormone and follicle-stimulating hormone secretion from the pituitary, the reproductive axis will remain quiescent.
41
Q

1B: Q13: skill 1: discrete

DNA ligase

Primase

Helicase

topisomerase

A
  1. DNA ligase: joins DNA fragments together usually in okazaki fragments
  2. primase sunthesized RNA primers needed to begin replication
  3. helicase unzips DNA
  4. tipoisomerase: prevents supercoiling
42
Q

2C: Q 16: skill 2

p38

A
  • p38 are known as “the gaurdians of the genome”, they work by inhibiting DNA damaged cells from continuing in interphase by stopping them at G1 so that DNA repair mechanisms can come in & try to fix the problem
  • so if p38 is overly activated, it will lead to growth arrest
  • passage indicated that p38 phosphorylation levels correlates with apoptotic signaling, which most likely leads to growth arrest
43
Q

2C: Q 17: skill 2: passage based

apoptosis/cancer

A
  • passage stated the cFLIP prevents the activation of capase-8, whcih would lessen the apoptotic response; inhibits apoptosis
    • inhibition of apoptosis is a characteristic of cancer cells
      • so if a cancerous cell is not undergoing apoptosis, that means that it will contiue to replicate & divide without ever dying. this will lead to cancer
44
Q

2A: Q 20: skill 1: passage based

the passage asked which membrane transport is directly affected by cardiac glycosides (is digoxin-inhibits N+K+ATPASE)

A
  • primary Active transport —-> ATPase
    • transporting substances against concentration gradient
    • Na/K+ ATPase pump is DIRECTLY coupled to active primary transport (taking substances against their concentration gradient). -ATPase/ Pumps are usually dead ringers for primary Active transport
    • Na/K+ ATPase is not directly coupled to secondary active transport. Creates an electrochemical gradient instead. secondary Active transport is usually characterized with antiport/synport

simple diffusion: passive transport (No ATP needed), no helper protein needed

faciliated diffusion: passive transport-no ATP needed, but requres a helper protein (a channel-ion/gated) protein or carrier such as uniporter, antiporter, symporter

45
Q

1B: Q21: Skill 4: graph based

A
46
Q

2A: Q 22: skill 1

Na+K+ATPase

A

3-2-1 NOKIA (where 3Na+ out, 2K+ in, using 1ATP).

47
Q

1B: Q 23: skill 4, graph based

pay attention to least, most questions

A
48
Q

1B: Q25: skill 1

histones/ HDACs/HATs

A
  • histone acetylation typically promotes transcription by modifying chromatin structure, HDACs would inhibit transcription by condensing chromatin structure (inc DNA coiling)
    • note: histone modification does not affect DNA replication, it impacts transcription of DNA

Histone acetyltransferases MASK the positive charge on the histones which in turn will make them LESS attracted to the negatively charged DNA (decreasing the charge repulsion between acetyl groups). Therefore the overall effect is *LOOSENING of the chromatin to increas*e transcription. (DEacetylases will basically do the opposite)

49
Q

1D: Q 26: skill 2

fasting/starvation & ketone bodies

A
  • fasting leads to glycogen breakdown and gluconeogenesis, then continued fasting leads to the production of ketone bodies by sustained fatty acid oxidation.
    • ​βOHB is a component of ketone bodies, so serum levels of βOHB are elevated when there is sustained fatty acid oxidation.
    • Sustained fatty acid oxidation (lipolysis) and ketogenesis are the major source of energy during starvation. And since the passage defines BHOB is a “major source of energy” during prolonged exercise and starvation, sustained fatty acid oxidation is the best answer

______________

  • in times of starvation, PPP is less active
  • Gluconeogenesis WOULD increase after onset of starvation. However, gluconeogenesis is not the biggest source of energy during starvation, and the passage states that BHOH is the “major source of energy during prolonged exercise or starvation.” REMEMBER: fatty acid oxidation (lipolysis) and ketogenesis are the highest source of energy during starvation.
  • Although the passage does state that BHOB is present during prolonged exercise - which triggers Cori Cycle (another name for lactic acid formation) - this does not explain why BHOB is also present during times of starvation.
50
Q

3A: Q 29: Skill 1

adrenal medulla, SNS

A
  • the adrenal medulla secretes epinephrine and norepinephrine in response to short-term stress. Reactions to short-term stress are mediated by the sympathetic branch of the autonomic nervous system.
    • ​both the pre and post ganglion of parasympathetic responds to Ach only whereas symapthetic’s post-ganglion responds to E/NE.
  • Adrenal Medulla: Catacholamines (Epi or Norepi) Adrenal Cortex: Mineralcorticoids (Aldosterone) and Glucocorticoids (Cortisol)
51
Q

1D: Q31: skill 1, discrete

which describes the chemical energy that is derived from the krebs cycle?

A
  • ATP, which directly supplies energy for many cellular processes (such as muscle contraction), & NADH, which supplies energy for the ETC
52
Q

1D: Q33: skill 1

lactate fermentation, regeneration of NAD+

A
  • Lactate fermrentaiton is triggered in response to anaerobic conditions to regenerate NAD+.
  • increased concentration of NADH inside the cell. For glycolysis to proceed, NADH must be converted back to NAD+ through lactic fermentation.
53
Q

1D: Q34: skill 1

phosphoglucose isomerase is in enzyme found where?

A
  • glycolysis
  • glycolysis activity increases during anaerobic metabolism, & phosphoglucose isomerase is a catalyst withing glycolysis
54
Q

2A: Q 36: skill 3, figure/passage based

tubilin/used as a control

A
  • Tubulin is a stable cytoskeletal protein and is often used as a control to compare the size of another protein during experiments.
55
Q

1B: Q 37: skill 3

resitriction enzyme, palindromic

sequence given: CCCCGGGC , which kind of restriction enzymes can recognize the HIF binding sequences? a restriction enzyme that has:

A
  • 4-base & 6-base recognition sequence
    • because only the CCCGGG within the HIF binding sequence is palindromic. Therefore, only a restriction enzyme that recognizes a four-base sequence or a six-base sequence can recognize this sequence within the HIF binding sequence.
  • restriction enzymes recognize palindrome sequences
56
Q

1D: Q38: skill 1

CAC Complexes I-IV

A
  • Complex I: NADH Reductase
  • Complex II: succinate dehydrogenase
    • Succinate feeds into Complex II of the ETC directly. Note: Complex II is the only enzyme that is part of both the citric acid cycle and the electron transport chain
  • Complex III: Cytochrome Reductase
  • Complex IV: Cytochrome Oxidase
57
Q

3B: Q42: skill 4, figure/passage based

A
58
Q

2A: Q 43, skill 2, passage based/figure

A
59
Q

1B: Q44: skill 4, figure based

A
60
Q

2B: Q45: skill 2, discrete

Viruses

Certain viruses contain RNA as their genetic material. One of the ways these RNA viruses replicate themselves is to:

A
  • code for or carry a transcriptase that copies viral RNA.
    • ​becuase RNA viruses require a type of transcriptase (reverse transcriptase) to replicate themselves.

_____________________

viruses will not going into an orangism with RNA as their genetic material, RNA -> Reverse transcriptase -> vDNA -> Integrase -> Integrates into host cell -> Transcription -> mRNA -> ETC…

+ sense RNA viruses rely on host ribosomes, they do not rely on host polymerase. + sense viruses basically deliver mRNA into the cell. These viruses have a large genome and one of the first proteins they code for is their own RNA replicase (RNA dependent polymerase) to replicate their genome.

61
Q

3B: Q46: skill 1, discrete

Kidney

Large amounts of protein are found in the urine of a patient. Based on this information, which portion of the nephron is most likely malfunctioning?

A
  • answer: glomerulus
    • healthy individuals, the structure of the glomerular capillaries prevents the entry of large molecules, such as proteins, into the filtrate.
    • Excreting proteins is bad business. We have the urea cycle for disposing of waste products from protein metabolism but we wouldnt want to be giving our proteins away by simply shoving them through the glomerulus. Similarily the spleen breaks down our dead RBC’s and we wouldnt want to be sending our RBC’s directly through the glomerulus without at least getting our heme group back first.

______________

Collecting duct: is associated with being the last line of control in terms of how much water will ultimately leave the kidney. If ADH is present the collecting duct will allow water to leave(the nephron/kidney) and therefore not be excreted. This section would not be associated with protein and red blood cell activity and if there were for some reason RBC’s making it this far they would continue on as the collecting duct does not have the machinery to facilitate such a transfer.

Distal tubule: uses active transport to play around with the ion’s and PH level of the filtrate it would not be expected to prevent protein or RBC excretion.

62
Q

2A: Q48: skill 1

tissues

Mucous secretions in the respiratory tract inhibit microbial infections. These secretions are produced by which of the following tissue types found in the lungs?

A
  • epithelial
    • because epithelial cells are specialized cells for secretion of muscus and are present in the respiratory tract
63
Q

3A: Q 49: skill 1

serotonin, pre & post receptors

A
  • passage said antidepressant drugs act to increase the amount of serotonin in the synaptic space.
    • Therefore, the action of serotonin on postsynaptic receptors would be terminated by the removal of serotonin from the synaptic space. This would occur by transport of released serotonin back into the presynaptic terminal, an action that SSRIs block
64
Q

3A: Q52: skill 2

SSRIs most likely relieve depression by increasing the:

A
  • answer: amount of time that serotonin is present in the synaptic space
    • passage states that SSRIs function to treat depression by blocking serotonin reuptake in the presynaptic neuron. Blocking the reuptake of serotonin into the presynaptic neurons would increase the amount of time that serotonin is available in the synaptic cleft to bind to postsynaptic neurons.
65
Q

3B: Q 53: skill 2

small intestines

what contributes to weight loss?

A
  • decrease in the surface area of the small intestine, leading to a decrease in nutrient absorption
    • the passage states that gluten, the factor that contributed to the boy’s weight loss, causes intestinal villi atrophy. Villi atrophy results in a decrease in the surface area of the small intestine, leading to a decrease in nutrient absorption.
      *
66
Q

3B: Q 56: skill 3

blood calcium

ways to increase blood calcium

A
  • answer: dec osteoblast activity & increase osteoclast activity
    • osteoblasts function to build and repair bone, osteoclasts break down bone. Therefore, to increase calcium levels in the blood, osteoblast activity should be decreased and osteoclast activity should be increased to release stored calcium from the bone to the bloodstream.
  • Key words: OsteoBlast: Build bones (absorbs calcium) Osteoclast: chews up/breakdown bones (release calcium to the blood stream)
67
Q

3B: Q59: skill 1

Kidney

Which series shows the order in which filtrate passes through the tubular regions of a nephron?

A
  • answer: Bowman’s capsule → proximal tubule → loop of Henle → distal tubule → collecting duct
68
Q

Which cells are the most highly proliferative in the body? give an example

A

epithelial cells

ex: gastrointestinal epithelial cells

69
Q

phosphodiester bond

https://www.reddit.com/r/Mcat/comments/b3rlq3/spoiler_aamc_fl2_bb_q3/

A

phosphodiester bonds link the 3ʹ carbon atom of one deoxyribose and the 5ʹ carbon atom of another deoxyribose within the DNA molecules

70
Q

phagocytosis

which cellular organelle is involved?

what do proteases break down, what about lysosymes?

A
  • macrophage ingests foreign material, the material initially becomes trapped in a phagosome.
  • The phagosome then fuses with a lysosome to form a phagolysosome. Inside the phagolysosome, enzymes digest the foreign object.
  • Remember: proteosomes break down proteins only
    • ​lysosomes break down fats, proteins, and carbs
71
Q

what inhibits PFK1? what kind of inhibition is this?

https://www.reddit.com/r/Mcat/comments/6nu8ig/spoiler_aamc_fl2_bb_q10/

A
  • inhibited by ATP
  • allosteric regulation & feedback inhibition
  • because ATP, the end product of glycolysis, downregulates through feedback inhibition the activity of phosphofructokinase-1 by binding to a regulatory site other than the active site of the enzyme (allosteric regulation).​
  • So this is a bit tricky because ATP is both a substrate and inhibitor of PFK-1. PFK has 2 sites for ATP, one active and one allosteric. The active site has a lower Kd than the inhibitory allosteric site. What does this mean? well at high concetrations of ATP (due to increase in glycolysis will bind to the high Kd low affinity allosteric inhibitory site and decrease PFK-1. But at normal ATP levels PFK-1 active site will be able to run normally.
72
Q

proteins with pI 9, what is their charge?

pI & pH

A
  • positively charged at physiological pH (7.4).
  • this means that the protein overall is basic.
    • NOTE: When the pH > pI, a protein has a net negative charge and when the pH < pI, a protein has a net positive charge
73
Q

secretory proteins are synthesized in the __________

smooth ER is responsible for the synthesis _______________

folding of secretory proteins also occurs in the ________________

A
  • rough ER
  • of carbohydrates, lipids, and steroid hormones, detoxification, and storage of Ca++ (sarcoplasmic reticulum of muscle cells).
  • rough ER
74
Q

Proteins that are translocated into the nucleus usually contain a _______________

A
  • nuclear localization sequence
  • the existence of a nuclear localization signal is what allows the protein to get to the nucleus
75
Q

Ubiquination

A

Ubiquitination targets a protein for degradation by a proteasome.

76
Q

transmembrane domains

A
  • cross the phospholipid bilayer. As a result, these domains are most likely to have a high proportion of hydrophobic residues.
77
Q

glyceraldehyde-3-phosphate dehydrogenase (GAPDH) catalyzes what reversible conversion?

A
  • glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate
  • review whats in blue in the excel sheet (Q20)
78
Q

posttranslational modification of proteins such as histone acetylation is analyzed by ___________

A

western blott

79
Q

RT-PCR

A
  • rt-pcr is a reverse transcription PCR from RNA to DNA to analyze the cDNA. This is good to use for analyzing gene expression.
80
Q

vasopressin

A

regulates the fusion of aquaporins with the apical membranes of the collecting duct epithelial cells.

81
Q

loop of henle

A
  • ascending limb is impermeable to water
82
Q

________are the myelin-forming cells in the peripheral nervous system

A
  • schwann cells
83
Q

______________macrophages in the CNS

A

microglia

84
Q

astrocytes

A

support cells in the CNS that also form the BBB

85
Q

Π = iMRT

A
  • memorize this osmotic pressure equation and be able to plug in #
  • where i is the van’t Hoff factor, M is the concentration of solute, R is the gas constant, and T is the temperature.
  • pay attention for what the Q is asking, in the practice test it asked for MOLARITY!!
  • The question is asking for MOLARITY not OSMOLARITY. Osmolarity would take into account the Vant Hoff factor whereas molarity does not. To solve for OSMOLARITY you would rearrange the equation to iM = (pi)/RT.
86
Q

PRION

A

a prion is an abnormally folded protein that induces a normally folded version of the protein to also adopt the abnormal structure, which is often deleterious.

87
Q

BLOOD FROM SMALL INTESTINS IS TRANSPORTED TO?

A
  • LIVER
  • which regulates nutrient distribution and removes toxins from the blood
  • Small intestine -> Hepatic portal vein -> Liver
88
Q

molecules made in the Citric Acid Cycle:

what is the mnemonic?

A

Officer, Can I Keep Selling Sex For Money Officer

89
Q

Meosis

In humans, eggs and sperm are most similar with respect to:

A
  • genome size!!
  • both eggs and sperm contain a haploid number of chromosomes and therefore they are most similar with respect to their genome size
90
Q

mesoderm

A

heart, kidney and muscle

91
Q

ectoderm

A

brain

92
Q

Lysozymes

A

Secretory lysosomes are classified as lysosomes, (so similarity is seen in hydrolytic enzymatic function) lysosomes are defined as membrane-bound organelles that contain hydrolytic enzymes activated by a low pH

93
Q

microtubules are cellular structures that originate from _____________

A

centrosomes

centrosome: an organelle near the nucleus of a cell which contains the centrioles (in animal cells) and from which the spindle fibers develop in cell division

94
Q

“x” binds to a promotor so what can we conclude about its regulation

A

regulation is expresses at transcriptional level

95
Q

sperm, produced in the______________ completes maturation and becomes motile in the ______________

A

seminiferous tubules of the testes

(also, testes produce testosterone)

epididymis (also stores sperm)

prostate gland secretes alkaline fluid that makes up most of semen, this produces lubrication & nutrition for sperm

96
Q

According to the cross-bridge model of muscle contraction, the muscles stiffen after death because ATP is unavailable to bind and directly release:

A
  • Answer: the myosin head from the actin filament.
    • ​this is also called rigoris mortis!
  • during normal muscle contraction, ATP is required to break the bonds between the actin filament and the myosin head. After death, no new ATP is generated, so the myosin head cannot be released from the actin filament, resulting in stiffening of muscles.
  • ATP is used to detach myosin from actin, so if no ATP avaiable to restart muscle contraction then the muscles stiffen, what gave it away to me was that it says “muscle stiffens” which i imagined they stay contracted and contraction happens when myosin binds to the actin and it shortens, causing contraction
97
Q

Reducing agents in SDS in 45 kDa protein will result in what? this is a homodiemer

A

adding a reducing agent would eliminate any disulfide bridges and allow the monomers to run separately–thus leading to a migration expected for the 22.5-kDa protein.

  • increasing gel running time–> this would allow two different proteins to separate more, but in this case there is only 1 so it doesn’t help
  • higher voltage would just melt the gel
  • sds is what coats the protein in even negative charge; if it is removed it won’t travel down to the positive end (bottom of the gel). it also denatures proteins so that they only travel based on molecular weight on the gel.
98
Q
A

MCAT AAMC Practice Exams/ QP/SB (4 decks)

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