How many different kinds of F1 gametes, F2 genotypes, and F2 phenotypes would be expected from the following crosses:
(a) AA X aa;
(b) AA BB X aa bb;
(c) AA BB CC X aa bb cc?
(d) What general formulas are suggested by these answers?
AaBbCC will produce 4 types of gametes which are as follows- ABC, AbC, aBC, abC. The number of gametes formed is decided by the number of heterozygous alleles present in the given genotype. 2^n is the formula used to find it out, where n=number of heterozygous alleles present in the genotype. Say for example, in the above genotype Aa & Bb are the 2 heterozygous alleles, so here n=2. Putting the values in the formula , we get 2^2=4. Hence 4 types of gametes are formed.
Comments
Leave a comment