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aamc cbt11 thread

Discussion in 'MCAT Study Question Q&A' started by sekistudent, Jan 2, 2011.

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  1. sekistudent

    sekistudent

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    SDN Members don't see this ad. (About Ads)
    Can we start an official aamc cbt11 thread. I took this test a few days ago and have some questions. Can I just start this thread? Or, is that sort of behavior for the more senior members.
  2. MilkIsGood

    MilkIsGood

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    havent taken it yet, but how does it compare to the other tests
  3. sekistudent

    sekistudent

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    I've also taken 3, 5, and 6. Its hard to say generally if this test is harder than the others. From a purely basic knowledge standpoint it compared evenly with the other tests. There weren't too many esoteric knowledge questions. And testing of concepts were pretty straight forward, I felt. They were basically looking for grasp of the general/popular concepts of a given topic. To that point however, I felt that the questions and passages were quite long and I had to read the question stems several times to clearly get what they were asking for. Also, there were more questions directly related to the passage.

    My strategy is to go over the questions first than hit the passage. This way I can get simple questions out of the way and also, get a sense of where I should be focusing. This proved to be difficult, and I'm glad I got to see this. The question stems were long and time consuming. So the whole time I was stressing whether I'm spending too much time looking at the questions first. Also, as I said there were more passage related questions and so I didnt even get a chance to answer many of them.

    The passages also took some more time analyzing. You definitely had to sit there and spend more time connecting the dots, if you know what I mean.

    My feeling is, they are gravitating further away from just knowledge based questions to analytical and interactive (passage) questions.
  4. Techmed07

    Techmed07

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    This sums up exactly how my mcat was when I took it.
  5. bmw e90

    bmw e90

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    Yeah, I found this exam to be quite difficult. Especially the Bio and the Verbal sections. I found a lot of the verbal answers required us to assume more than normal. When questions asked what does this imply?, i felt the answers were farfetched imo. And the Bio section was tough. They way they worded the question stems didnt help either.
  6. plzNOCarribbean

    plzNOCarribbean

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    yeah i agree with this completely. Bio is usually my best section and i did pretty poorly here, partly because of dumb mistakes due to the time constraints, as mentioned above.

    I really do like the idea though of just taking a few seconds to glance over the questions prior to reading the passage. I thought the PS wasn't bad, just much more calculation heavy.
  7. gunj122

    gunj122

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    i felt that the verbal was difficult. however, while i was taking it, i didnt think it was too bad but when i got my score back, that was a different story. also, i think i was just distracted from personal issues that arose right before i sat down to take it. but anyway, the sciences were alright. PS was straightforward for the most part. BS got a little out of hand in a few bio passages with the jumbling of acronyms like VSV-EGP, or whatever it was. Honestly, the fundamental concepts don't change, so what theyre testing is still easy. but its just presented in a convoluted fashion.

    I take my test on Saturday =O i'm just crossing my fingers for verbal to swing my way and at the most, 2 orgo passages. the fewer the better haha.
  8. agill786

    agill786

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    For BS Item 34, I don't see why cytoseine tRNA would not be related to the cp450b, doesn't the passage say that the protein binds to cytoseine somewhere?
  9. JeetKuneDo

    JeetKuneDo

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    Is this question 124?
  10. agill786

    agill786

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    I took the bio portion by itself so it is passage 4 last question.
  11. nebratu

    nebratu New Member

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    Posting in here cause this is the only CBT 11 thread I found.

    For question 31, is |Wg| going to be greater than |Wa| until terminal velocity is reached? Will the abs values be equal once terminal velocity is reached?
  12. attixx

    attixx

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    I'm having trouble understanding this as well. They say during the first 600m, and at 600m terminal velocity is reached. At that point, I feel like the work of air resistance should be equal and opposite to the work of gravity since at that point there is no net force. AAMC does not seem to think so.
  13. Catburr

    Catburr

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    I was reviewing this test last night, and I feel like their explanation is a little confusing too. But here's what I got from it.

    It's a setup where something that is released from rest, and reaches terminal velocity after falling 600m. The question asks about work by gravity and work by air resistance DURING that first 600m. My intuition was, velocity is down, so work by gravity is positive, and work by air resistance (going against velocity) is negative. That settles the first part.

    Now what about the absolute values of work? That's the weird part, or at least it was for me. And the explanation doesn't even seem to address it, just says "here's how you calculate work by gravity, and oh, air resistance does negative work, therefore Wgrav > Wair."

    I think the key is, the falling object started from REST, and accelerated down until terminal velocity was reached. At that moment, of distance traveled = 600m, he's falling at terminal velocity. So gravity must have done more work, since we started at rest, and now he still has kinetic energy pointing down.

    Kind of like photoelectric effect, where a really high energy photon gives the electron it kicks off some extra energy to travel away with? Bad analogy? Probably.

    I'd love if someone else has a better analysis. CoughCommissionerRaboCough.
  14. Rabolisk

    Rabolisk

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    I see that I'm finally getting the recognition I deserve!! (sarcasm)

    I agree with you on everything pretty much. The magnitude of the work done by gravity had to be greater than the magnitude of the work done by air resistance because kinetic energy increased from 0 to some number. Since air resistance does negative work, it wants to decrease kinetic energy, while gravity wants to increase kinetic energy. If |Wa| > |Wg|, then Wg + Wa < 0, which means that the object's kinetic energy decreased from 0 to.. some negative number. Clearly impossible.

    Another way to look at it is by using W = Fd. Since the object accelerates downward, the force due to gravity must be greater than the upward force due to air resistance. The gravitational force is constant, and the air resistance increases until it is equal and opposite to gravity, which is when terminal velocity is reached. But since Fa < mg for at least some time period, |Wa| < |Wg|.

    I also agree that the photoelectric effect is just a bad analogy. You are right so many counts today!
  15. attixx

    attixx

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    I see how you are correct. But tell me why this thinking is wrong:

    for 600 feet, there is a downward work = F*d = mg600.

    Since air resistance also acts for the same distance, d is the same. Also, I thought force of air resistance is the same since at terminal velocity, F grav = -F air resistance.

    I know there is a flaw in that argument. Is it because gravity acts at the full extent of its force the entire 600 feet, and air resistance starts at ~0 and increases to its value at 600 feet?
  16. Rabolisk

    Rabolisk

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    Yes. Work is really the line integral of force over a distance. So if the magnitude of force is not the same for the entire distance or path, then work is not equal. Of course angle matters too, but in this case, they are clearly 0 or 180 degrees. |Wg| and |Wa| after the object has reached terminal velocity is equal.
  17. Catburr

    Catburr

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    I like this analysis much better than my original one. Makes so much sense, thanks a lot.

    This made me laugh. I figured as much. : ) But seriously, the skydiver is the electron, the plane is the atom it was originally in, the distance between rest and terminal velocity is the process of leaving the atom, and the terminal velocity is the extra energy the electron has...ok I'll stop. This is why my day job isn't Kaplan instructor or something. Though from some of the Kaplan instructors I've met, I probably wouldn't be much worse, but that's a whole other story.

    That thinking is wrong only for the part of the fall that the question's asking about. You're thinking of the moment terminal velocity is reached, and the fall afterward. For those times, force by gravity and force by air res. is equal, because there is no net force, just constant velocity. For a random 600m portion of the fall AFTER terminal velocity is reached, absolute value of work by gravity and by air resistance would be equal.

    Yup. Like Rabo said, air resistance increases until it is equal and opposite gravity. While air resistance is trying to catch up, the jumper is accelerating from rest to terminal velocity.

    EDIT: Derp, too slow.
  18. dlouis

    dlouis

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    I'm taking this test on Wednesday. I'm taking the real test on Saturday 5/7 so we'll see how close my scores are :D My friend told me that a kid he studied with took it, got a 24, took the real thing and got a 24. I'm kind of glad that the test may be difficult because the real thing sure as hell will be.
  19. WhatMD88

    WhatMD88

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    Hey guys,

    I had a question on the PS (question#14) about the angular momentum quantum number for Mg2+ after losing its two electrons?

    I selected 2 because it goes from 3s2 to a 2p6 when it forms a cation. But its apparently......spoiler alert......zero.
  20. Catburr

    Catburr

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    That question isn't asking about the cation's angular momentum quantum number. It says, "What is the angular momentum quantum number (l) for the orbital FROM WHICH a Mg atom loses two electrons to form Mg2+?"

    So the answer is 0, because the orbital that loses the electrons is s. Also, the p orbital angular momentum quantum number is 1, not 2.

    Angular momentum quantum numbers:
    s = 0
    p = 1
    d = 2
    f = 3
  21. movax

    movax

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    Just took this one, maybe it was my nerves or the fact I took right after CBT9, but score went down from 39 on 9 to 33. I've been practicing with the full 77-question/section paper tests from 4-9, so I don't know if it was the abrupt transition to a computer or what, or just the steep curve. Missing 4 questions in PS on 9 gave me a 14, and that was with 77, not 52!

    Going through solutions now, but not the outlier I wanted 4 days before my test.
  22. cgk

    cgk Member

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    My friend just took AAMC test # 11 and his verbal score really plumetted. Did anyone else find this test's verbal section exceptionally difficult relative to the other AAMC CBT verbal sections?
  23. dlouis

    dlouis

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    AAMC 11: PS-8, VR-10, BS-11

    5/7/11: PS - 10, VR-10, BS-11

    I knew the 8 was a fluke, I went in to the test center confident, stayed confident the entire test, and scored almost exactly my #11 score. (luckily 2 points higher!)
  24. movax

    movax

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    I found it much easier actually, maybe because it was only 40 questions instead of 65, and there was no obnoxious Picasso or otherwise art passage.

    Really hoping this one isn't a true predictor of my score...how soon before the test can you change your date again?

    e: I also found the Ebola passage particularly hard to comprehend, especially that graph...maybe my brain was just fried at that point. Missed 4 alone there.

    What does each heading mean? 'CatB-/-', does that mean no CatB? So 'CatL+/+' means presence of CatL, 'CatL-/-' means abscence of CatL? So for 'CatB-/-,CatL+/+', when CatB is an "introduced" gene, the cell now has both CatL and CatB?
    Last edited: Jun 12, 2011
  25. turkeybean

    turkeybean

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    I actually scored the highest on this exam, especially the PS section. However, I thought the BS section was quite impossible, and some questions were very unclear.

    For BS #129 (the Sn2 rxn problem), how could you tell what the 'D' is supposed to stand for? Without knowing that, how would you assign priorities?

    Then, on #132, the question states "If Experiment 2..." but I do not see where in the passage an "experiment 2" is mentioned. I ended up reading through the whole thing.

    Am I missing something here? I lost precious time over these trivial points and thought at least this passage was very poorly designed. I hope I don't run into anything like this on the real thing.
  26. movax

    movax

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    D is Deuterium, aka heavy water, I think this is expected knowledge for the MCAT.
  27. turkeybean

    turkeybean

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    OK - thanks! I actually did consider it as heavy hydrogen to answer the question but worried that I might be missing some other important link -- I hadn't seen many of these structures before (shoddy chem fundamentals) and didn't think they would use that just to make a point.
  28. ThirdEye

    ThirdEye

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    Yes this is how it works, but I couldn't figure it out during the test either. Going back to it now still makes my head hurt. If they throw something this dense on the real deal it better be experimental :xf:

    BTW, why do all of the AAMC explanations suck? Their explanations suck for all of the simple problems too. Who writes this nonsense?
  29. ThirdEye

    ThirdEye

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    They meant Figure 2. I was tripped up on this too but not for too long. You have to keep in mind that the people who write these tests have a greater interest in making us miss questions than they do in making it comprehensible.
  30. Axon hillock

    Axon hillock

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    I thought the verbal was more difficult than the other exams as well. My score didn't plummet b/c I think the conversion was slightly more favorable but it definitely seemed tougher.
  31. lenomaly

    lenomaly

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    Hi, this is my first post on SDN. For PS 27, how do you get "P=Fv=-mgVt" from passage? Also, not really getting PS 28 either... Thanks in advance!
  32. Chowdder

    Chowdder

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    PS 27
    This was kinda tricky for me too, because the alternative power formula is not something that I'm used to so it took me a while to remember. But Power is also equals to force times velocity, hence P=FV.

    The question is asking what is the power of air resistance after reaching terminal velocity. Therefore, you know right off the bat that VT must be in the equation. Because at terminal velocity, the object is no longer accelerating, then we know that the downward force is equal to the upward force. the downward force is mg, the weight of the object (gravitational force). So the upward force must be -mg.

    So P=FV which in this case F = -mg and v = VT.

    TIP: If you're really stuck on this type of question and don't know where to apply the formula, I'd recommend playing around with the units. Units of Power is W = J/s = kg*m^2/s^3. If you look at all the answers, only A, P = -mgVT and C, P = -1/2mVT^3/h has the correct unit. You can eliminate C because Power at terminal velocity should not depend on h.

    PS28
    You should know the relations of distance versus time in a uniform accelerated motion. Where distance is proportional to time squared. However, once terminal velocity is reached, acceleration is 0, which means that distance is now proportional to only time. Thus the graph becomes linear. So the first part of the graph should have a exponential increase while the later half should be a linear increase.

    A is wrong because it implies that acceleration is 0
    C is wrong because it implies that acceleration is negative (decreasing velocity)
    D is wrong because it implies that you have positive acceleration, then 0, then negative acceleration.
    Last edited: Jul 28, 2011
  33. collegestud2013

    collegestud2013

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    Which of the following statements best explains why air resistance is ignored when a compact object falls a very short distance?

    A) The object's mechanical energy is constant.
    B) The speed of the object remains small.
    C) The area presented to the air negligible.
    D) Gravity does negligible work in this situation.

    The answer is B), but I personally thought C) was also justifiable since the object is "compact" so its "area presented to the air is negligible." And since the net force on the object as it falls is:

    F = mg - b(v^2) = m(dv/dt)

    where bv^2 is the term equal to air resistance, in which b is a constant that's positively related to the "area presented by the object perpendicular to the motion through the air." So since both b and v are small, wouldn't BOTH contribute the fact that air resistance can be ignored. The two reasons I could think of why B) is a better choice is that the air resistance is proportional to the square of speed while it's only proportional to b, so speed would have greater impact on air resistance. The other is that they were implying which is effect is relevant only at a very short distance but NOT long distance, and since b is is a constant while v is increasing with distance, b would still contribute the decreased air resistance while v would become to large at longer distances. Kinda frustrates me since this is the only question I missed on PS for this exam.
    Last edited: Aug 1, 2011
  34. salim271

    salim271 Patience is tough. :/

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    Wasnt the only one I missed lol but I had the same exact thinking as you. Some of the justifications they give seem weak to me.
  35. dmplz707

    dmplz707

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    Hey guys I have question on PS # 30. I the magnitude of the air resistance equals the weight of the parachutist, then how will the parachutist fall since the sum of all the forces in the vertical direction would be zero?

    What is the magnitude of the air resistance force while the parachutist traveling at vT?

    A 40 N

    B 75 N

    C 750 N
    The air resistance on the parachutist at terminal velocity equals the weight of the parachutist which is given by F = m g = 75 kg × 10 m/s2 = 750 N. This is option C.

    D 3000 N
  36. collegestud2013

    collegestud2013

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    It would fall at constant velocity from gravitational PE that has been converted to KE since there is no net force on the person (remember no net force does not say anything about the object's actual speed, just that it's speed is not changing, or a = dv/dt = 0).
  37. mitchlucker

    mitchlucker

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    Can someone explain to me what CatB -/- means, or CatL +/+ means? Or something like CatB +/-? I'm not understanding this notation.
    Last edited: Aug 1, 2011
  38. collegestud2013

    collegestud2013

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    The "+" indicates an allele in the gene coding for the CatL protein that is ABLE to make the protein, so it is the functioning allele. The "-" is a mutant allele of the gene that is non-functional, and thus unable to properly code for the protein. Thus, the cells that were initially CatB -/- and CatL +/+ are able to synthesize the CatL protein but not the Cat B protein.

    And since the presence of BOTH proteins AND the EGP viral glycoprotein are required for viral infection, the virus would not be able to infect mutant forms of the cell unless the necessary functioning versions of the gene(s) that were -/- were exegenously inserted.
  39. mitchlucker

    mitchlucker

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    Thanks. Passage seems a lot clearer with this explanation.

    Still unclear as to why bother with the " / " and the dual +/+ or -/- from your explanation.
  40. icecoldstar

    icecoldstar

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    This "+/+" notation comes from scientific papers (research labs). For example, they would say a mouse is +/+ for long tail. That means that mouse is homozygous dominant for long tail trait. If they say +/- for long tail, that means that mouse is heterozygous for long tail. and if they say mouse is -/-, that means it is homozygous for short tail. It's just a type of notation they use. so just apply this same logic for CatL and CatB.
  41. mitchlucker

    mitchlucker

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    Gotcha, thanks.
  42. dmplz707

    dmplz707

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    Hey guys I have a question on # 107 in the BS section. I was wondering if the concept being tested here is in relation to the inactivation of an X chromosome in a female somatic cell (Barr Body). Thanks.

    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 Inactivation of one R chromosome in males
    The question asks the examinee to apply knowledge about sex chromosome inactivation to a system in which the male is homozygous for the sex chromosomes and the female is heterozygous. By analogy with the mammalian system in which the production of genetic products (mRNA and protein, for instance) from the two X chromosomes in the female is reduced by randomly inactivating one of the X chromosomes, the conclusion can be reached that if the male is homozygous (RR in the system described), inactivation of one R chromosome in males would achieve the appropriate reduction in output of genetic products. Thus, A is the best answer.

    B Doubling transcription from the S chromosome in females

    C Inactivation of the R chromosome in females

    D Doubling transcription from the R chromosomes in males
  43. collegestud2013

    collegestud2013

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    They're specifically testing your knowledge of dosage compensation and how/why it occurs. In humans, the X chromosome contains more genes than the Y choromosome, so females (XX) would have more genes than males (XY) if one of the female's X chromosomes weren't inactivated. In this specific question though they want you to realize that it doesn't have to occur in females of this species just because that's what happens in humans, but that the sex with the homozygous pair of sex chromosomes must inactivate one of them to match the same number of genes as the sex with a hetereozygous pair.
  44. TheRealAngeleno

    TheRealAngeleno

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    Sup guys I have a question on # 121 of the BS section. My question is in response to the bolded statement below. Could someone please explain how decreased blood osmolality is indicative of high levels of hydration? Thanks a lot guys!


    Which of the following physiological conditions would be LEAST likely to induce thirst?

    A. Dry pharynx

    B. Decreased blood volume

    C. Decreased blood pressure

    D. Decreased blood osmolality
    This question asks the examinee to identify the physiological condition LEAST likely to induce thirst. Of the options listed, decreased blood osmolality is the option most indicative of high levels of hydration. Thus, D is the best answer.
  45. collegestud2013

    collegestud2013

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    remember osmotic pressure of of a solution = iMRT. So if blood osmolality is low, that means it is more dilute (smaller M) and contains less solutes (like albumin proteins and Na+), and since less solute for a given volume implies more solvent (in this case the blood plamsa itself), the high concentration of blood plasma indicates high levels of hydration.
  46. TheRealAngeleno

    TheRealAngeleno

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    Does anyone know what the significance is of the introduced bacterial gene in Figure 1 of Passage 7 in BS section? Is it just a control?
  47. TheRealAngeleno

    TheRealAngeleno

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    Thanks a lot!
  48. mitchlucker

    mitchlucker

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    There is no figure 1, but I'm assuming you're talking about question 1.

    There is no significance. It was a "discrete" question basically, with some relevance to the passage.

    Just asking you what bacterial transformation is.

    If you're referring to the passage as figure 1, it was just a piece of information. Didn't have anything to do with the questions.

    Unless you didn't know what transformation was, in which you could conclude that to alter the expression of a protective capsule, you need to have a nucleic acid change.
  49. TheRealAngeleno

    TheRealAngeleno

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    Ah sorry I meant the figure in passage 6. My bad.
  50. PingPongPro

    PingPongPro

    Joined:
    Jan 28, 2010
    Messages:
    584
    Status:
    Medical Student
    SDN 2+ Year Member
    .
    Last edited: Sep 6, 2011

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