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So I have been working through the drills in the 26th edition of AP Biology Prep by The Princeton Review for fun when I came across this question in the Chapter 12 drill that I'm not really sure about my answer to:

  1. Given the cross $AaBbCc\times AaBbCc$, what is the probability of having an $AABbCC$ offspring?

    (A) $\dfrac14$

    (B) $\dfrac18$

    (C) $\dfrac1{16}$

    (D) $\dfrac1{32}$

Here's my thinking:

I can draw up 3 Punnett squares to calculate the probability of having an $AA$, then a $Bb$, and finally a $CC$ offspring, as then due to the law of probability, I can then multiply these probabilities together to get my final answer, right?

So we have the three Punnett squares: $$ \begin{array}{c|c|c|} & A & a \\ \hline A & AA & Aa \\ \hline a & Aa & aa \\ \hline \end{array}\\ \\ \begin{array}{c|c|c|} & B & b \\ \hline B & BB & Bb \\ \hline b & Bb & bb \\ \hline \end{array}\\ \\ \begin{array}{c|c|c|} & C & c \\ \hline C & CC & Cc \\ \hline c & Cc & cc \\ \hline \end{array} $$

So from our Punnett squares, we have a 25% chance of the offspring having the AA gene, 25%+25%=50% chance of the offspring having the Bb gene, and then a 25% chance of the offspring having the CC gene.

So, due to these observations, the probability of this cross resulting in an $AABbCC$ offspring should be $1/4\times1/2\times1/4=1/32$, and (D) is the correct answer.

My question is: Is my thinking correct, or is my answer wrong, and why if so?

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    $\begingroup$ Yes, it is correct. $\endgroup$
    – Domen
    Commented Apr 6 at 22:02
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    $\begingroup$ In the case of no linkage, yes. $\endgroup$
    – Maximilian Press
    Commented Apr 11 at 16:58
  • $\begingroup$ @MaximilianPress What do you mean by linkage? I've never heard of it. $\endgroup$
    – CrSb0001
    Commented Apr 12 at 17:54
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    $\begingroup$ Linkage is the non-independence of transmission of alleles at different loci. It is traditionally understood to mean when two loci (e.g. the A locus, or B or C loci, from your example) are physically close to each other on the chromosome. This leads to distortions because it is unlikely that recombination will reassort alleles from those 2 loci. There are also other phenomena that can lead to linkage. I strongly recommend any genetics text to learn more. $\endgroup$
    – Maximilian Press
    Commented Apr 12 at 18:39

1 Answer 1

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Your thinking is indeed correct. Just from probability, if you know the odds of 3 separate events, multiplying them together will give you the % chance of all 3 occurring. Putting this whole thing into a single punnet square is also possible which gives us a 2/64 chance, or 1/32 as the answer choice says. If I were taking an exam I would use the same thinking as you. enter image description here

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    $\begingroup$ Sorry about the late comment @Snow, but welcome to Bio.SE! This is a great first contribution, and I hope you enjoy your stay here! I actually would like to provide some feedback about the giant Punnett square: While I do find the idea interesting, wouldn't this be super inefficient to use when taking a test? Or is that why you said that you'd use the same thinking as me during a test? $\endgroup$
    – CrSb0001
    Commented May 2 at 14:10
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    $\begingroup$ Hello @CrSb0001, thanks for your friendly remarks. Yes apologies if I was not clear enough but when I said I would use the same thinking as you in an exam I mean precisely that I would just multiply the separate odds together to get the total. Even drawing a 2x2 square in an exam is not particularly useful unless in strange cases such as X or Y chromosome problems, where they can be very helpful. My 8x8 monstrosity was just to help give you the full picture and to act as proof of your idea. I would never use that in an exam setting. Good luck with the AP biology prep! $\endgroup$
    – Snow
    Commented May 3 at 20:42
  • $\begingroup$ Thanks, although I'm just doing it for fun lol $\endgroup$
    – CrSb0001
    Commented May 4 at 16:45

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