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.

Question

Darnell Ramsey

Biology

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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Answers (1)

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Chris Haynes · Answer 1

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.