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28 March, 03:27

You set up the following cross: Aa:Bb:Cc:Dd x aa:Bb:Cc:dd What proportion of the offspring should have the genotype Aa:bb:CC:dd? Assume all alleles represented by capital letters are dominant and all genes are independently assorting.

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  1. 28 March, 03:35
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    Answer: The correct answer is 1/64

    Explanation:

    Given cross is AaBbCcDd x aaBbCcdd

    Now, if we consider the cross Aa x aa, the probability of having Aa is 2/4

    Similarly, for Bb x Bb, the probability of having bb is 1/4

    Cc x Cc, probability of having CC is 1/4

    Dd x dd, probability of having dd is 2/4

    Thus, the probability or proportion of the offspring to have genotype AabbCCdd = 2/4 x 1/4 x 1/4 x 2/4 = 4/256 = 1/64

    Thus, only one out of 64 offspring is expected to have this genotype.

    Alternatively, one can draw the Punnett square of AaBbCcDd x aaBbCcdd

    Total number of possible genotype combination which can be formed from above cross can be found using the formula 2ⁿ where n is the number of heterozygous pairs.

    In AaBbCcDd, there are 4 heterzygous pair and in aaBbCcdd, there are 2 heterozygous pairs.

    Thus, n = 6

    2⁶ = 64

    16 type of gametes will be formed by AaBbCcDd and 4 types of gametes will be formed by aaBbCcdd.
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