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genetics question

Discussion in 'DAT Discussions' started by allstardentist, Jul 23, 2006.

  1. allstardentist

    allstardentist All-Star
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    Laboratory mice are to be classified based on genes A,
    B, and C. How many genetically distinct offspring can
    be produced from a cross of an AaBbCc individual with
    an AaBBCc individual?
    A. 12
    B. 16
    C. 32
    D. 48
    E. 64

    I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
     
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  3. 714guy

    714guy counting down till 5/2012
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    For AaBbCc you have 3 heterozygous genes so 2^3 = 8 ways, and AaBBCc there are only 2 heterrozygous genes so 2^2 = 4 way. 4*8 = 32
     
  4. jorUW

    jorUW Senior Member
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    You have to multiply by two b/c they can be male OR female. Tricky Question
     
  5. allstardentist

    allstardentist All-Star
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    i dunno this is what i thought..
    possible combination for A: AA Aa aa
    possible comb. for B: BB Bb
    possible comb. for C: CC Cc cc
    so i just multiplied 3x2x3 since we can prolly assum independent assortment. what is wrong with my logic?
     
  6. jorUW

    jorUW Senior Member
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    I dont know if this is the correct way to do it b/c this math counts the inheritance of 'A' mom 'a' dad Vs 'a' mom and 'A' dad as two distinctly different genotypes. I could be wrong, but i think it's b/c you have to account for female and male offspring like I said above.
     
  7. allstardentist

    allstardentist All-Star
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    multiplying by 2 will give me 36 not 32
     
  8. allstardentist

    allstardentist All-Star
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    shouldnt 'A' from mom and 'a' from dad be same as 'a' from mom and 'A' from dad? because they will both yield geneticallysame offspring.
     
  9. jorUW

    jorUW Senior Member
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    I think you're right, but you miss the last step. In one of my genetics classes my prof liked to add little statements like whats the probability of an offspring with genotype blah blah blah THAT IS FEMALE (except she wouldn't highlight the tricky part of the Q). A lot of people forget to factor this in. You get so focused on the polygenic trait that you miss the easy part.
     
  10. jorUW

    jorUW Senior Member
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    hmmm, my bad :D I'll look at it again.
     
  11. mlle

    mlle Senior Member
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    I agree with 714guy.

    Allstar, if you do it your way you aren't accounting for all the ways each of those could combine with eachother.

    If you had a simpler question like how many different gametes might form from an AaBbCc parent?

    Answer: 8 (draw it out if you need to)

    or how many could form from a AaBBCc parent?

    Answer: 4

    8*4=32.

    see?
     
  12. jorUW

    jorUW Senior Member
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    Maybe 714guy is right. I follow your logic, though that Aa vs aA would be the same.
     
  13. R.L.H

    R.L.H LoVe mY liFE
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    Totally agree! Same logic! 8*4=32 :thumbup:
     
  14. Uracil

    Uracil Member
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    In any problem, 2 ^ # of heterozygotes = the number of genetically distinct offspring that can be produced.

    In this case, you have a cross of AaBbCc X AaBBCc

    Parent A (AaBbCc) is heterozygous for the genes A, B and C. So, the number of heterozygotes = 3

    Parent B (AaBBCc) is heterozygous for the genes A and C only. Then here we have that the number of heterozygotes = 2

    2 + 3 = 5

    2^5 = 32



     
  15. jorUW

    jorUW Senior Member
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    So, even though there are only 18 possible genotypes, whether a given gamet is inherited by mom or dad makes the offspring distinct? I mean if the Q asked how many possible gentoypes or phenotypes would it be 18?

    Possible Gentotypes/phenotypes (assume incomplete dominance) w/o respect to how inherited, agree?

    1.AABBCC 2.AABBCc 3.AABBcc 4.AABbCC 5.AABbCc 6.AABbcc
    7.AaBBCC 8.AaBBCc 9.AaBBcc 10.AaBbCC 11.AaBbCc 12.AaBbcc
    13.aaBBCC 14.aaBBCc 15.aaBBcc 16.aaBbCC 17.aaBbCc 18.aaBbcc

    Edit Anyway, I understand the math. I've been out of genetics way too long :D
     
  16. allstardentist

    allstardentist All-Star
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    so we need to assume the above to be genetically distinct? That doesnt make sense to me.
     
  17. teefRcool

    teefRcool Senior Member
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    Yup i agree thats how I would do it
     
  18. allstardentist

    allstardentist All-Star
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    that problem is different cuz you don't have to worry about redundancy since it is asking for the possible different types of HAPLOID cell
     
  19. Uracil

    Uracil Member
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    There are 18 possible genotypes when the parents are crossed, you are correct.

    The problem is asking about the different gametes that can possibly be formed treating genes A, B, and C individually. If the problem was asking you about the different genotypes, yes it would be 18 :thumbup:



     
  20. allstardentist

    allstardentist All-Star
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    it is asking for different genotype(=offspring) not different types of gamets
     
  21. rchuloholla

    rchuloholla Probationary Status
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    oh i had a question just like this on the real DAT!
     
  22. Uracil

    Uracil Member
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    yeah that's right, I was confusing myself!! :D
     
  23. allstardentist

    allstardentist All-Star
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    uh oh. i better figure this out
     
  24. tinman831

    tinman831 ¯\_(ツ)_/¯
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    I think everyone is reading too much into the question. There are only 18 possible genotypes for the offspring. They are not talking about possible haploid gametes, just the genotypes of the full diploid organism. The problem didn't indicate that the male/female chromosomes were involved in this either. The only possible genotypes using the alleles that was indicated in the question is given in jorUW's post.

    OP where did you get this question from?
     
  25. dat_student

    dat_student Junior Member
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    AaBbCc * AaBBCc

    (4 gametes, 3 distinct gametes)
    .....A.....a
    A...AA...Aa
    a...aA...aa

    (2 gametes, 2 distinct gametes)
    .....B.....b
    B...BB...Bb

    (4 gametes, 3 distinct gametes)
    .....C.....c
    C...CC...Cc
    c...cC...cc

    # of possible gametes:
    4 * 2 * 4 (i.e. 2^2 * 2^1 * 2^2) = 32 genetically distinct offspring based on A, B, C and other genes.

    or
    3 * 2 * 3 = 18 distinct just based on A, B & C
     
  26. allstardentist

    allstardentist All-Star
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    from kaplan natural science worskshop a #29
     
  27. allstardentist

    allstardentist All-Star
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    hey DAT student, so u think the answer is correct?
     
  28. dat_student

    dat_student Junior Member
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    I think the answer is correct. c from mother may be linked with gene Z but c from father may be linked with lower case z. cZ is not the same as cz. So genetically 32 & phenotypically it can be 18 (if A, B & C are co-dominant).
     
  29. tissy

    U.S. Public Health Service 10+ Year Member

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    Second that..
     
  30. biocmp

    biocmp I'm a computer
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    This is a simpler question than what it sounds at first. Look at each genetic set seperately.

    Aa X Aa will give you 3 combinations. AA, Aa, aa
    BB X Bb will give you 2 combinations. BB, Bb
    Cc X Cc will give you 3 combinations. CC, Cc, cc

    This gives you only 18 different possibilities. There is no discriminate difference between inheriting 'a' from mom or 'a' from dad. So, unless the question was worded different, this is the only answer. It did not ask for more than this....

    If you do no agree, please explain. preferably with punant squares and good explanations. I want to make sure i understand this
     
  31. dat_student

    dat_student Junior Member
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    example:

    Chromosome 1 (mother) ------c-----Z (upper case)-------
    Chromosome 1 (father) -------c-----z (lower case)-------

    Are you saying the two chromosomes above are duplicates?
     
  32. biocmp

    biocmp I'm a computer
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    what does your Z stand for? I am not saying anything about sex chromosomes. Or perhaps I'm really tired. The question implies nothing about sex chromosomes. It simply gives you two genetic sets and asks the total number of recombinations
     
  33. dat_student

    dat_student Junior Member
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    Z is an allele. Chromosome 1 is not a sex chromosome :) The question says the mice are classified based on genes A, B and C. Those genes can be linked to other genes (e.g. Z).
     
  34. jorUW

    jorUW Senior Member
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    If you look at it this way than the answer is correct (32). I just think it's worded badly. Maybe you're just supposed to assume. Personally it is difficult for me to deduce that an A from mom is different than an A from dad based on how the question is worded. Even so, i understand in real life it would be.
     
  35. allstardentist

    allstardentist All-Star
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    If we follow DAT students logic, shouldnt we consider the two Bs in BB as distinct gametes? So it should be 4x4x4? It doesnt make much sense to me
     
  36. midwestboy

    midwestboy Member
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    2^5 for number of heterozygous traits. definitely right!
     
  37. biocmp

    biocmp I'm a computer
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    That's kind of disheartening if they are going to have questions worded like that thru the whole thing. I will get prepared enough for my own sake, but i would hope they would be a little more specific in their wording.

    I'm trying to figure out how I shoud deduce what they want?? :oops:


    So, DAT student, is that how I should aproach every genetics question on the DAT? I won't be taking it until next summer, but I am busy going through the book a few times and making sure I understand possible questions.
     

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