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.
1 answer:
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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