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Discussion in 'MCAT Study Question Q&A' started by annospree, Dec 11, 2008.

  1. annospree

    2+ Year Member

    Aug 12, 2008
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    What is the probability that the offspring of this cross (AaBbcc x AabbCC) will show at least 2 dominant traits?
  2. ksmi117

    ksmi117 GEAUX TIGERS!!!
    Physician Moderator Emeritus Lifetime Donor Verified Account 7+ Year Member

    Mar 15, 2008
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    You can go about this problem two ways.

    First, you can take locus separately. Aa x Aa gives 3/4 dominant (AA and Aa), Bb x bb gives 1/2 dominant (Aa), and cc x CC gives all dominant (Cc). You need to find the probability that at least two of these are true. We see that C will be dominant for all possible combinations, so all we need to find is the possibility that either A or B will be dominant. We have these cases 1. A dominant, b recessive 2. B dominant, a recessive 3. A, B dominant.

    The probability of dom. A and rec. b is 3/4 x 1/2 = 3/8
    The probability of rec. a and dom. B is 1/4 x 1/2 = 1/8
    The probability of dom. A and dom. B is 3/4 x 1/2 = 3/8

    So the total probability is the sum of the three = 7/8.

    The other way is to do the trihybrid cross. It may seem daunting at first, but this situation actually turns out quite simple. First, we need to find all the possible haploid genotypes. For the first part of the cross (AaBbcc), we can have ABc, Abc, aBc, or abc. For the second (AabbCC), we can have AbC or abC. Just do the cross and you find:

    x . . . ABc . . . Abc . . . aBc . . . abc
    AbC AABbCc AAbbCc AaBbCc AabbCc
    abC AaBbCc AabbCc aaBbCc aabbCc

    All but the lower right corner has two dominant traits which is 7/8.

    Hope this helps!
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    #2 ksmi117, Dec 11, 2008
    Last edited: Dec 11, 2008

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