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.

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.