PLEASE ANSWER THE QUESTION COMPLETELY Mice of genotypes AABB

PLEASE ANSWER THE QUESTION COMPLETELY! Mice of genotypes AABBCCDDSS and aabbccddss are crossed. The F1 offspring are then intercrossed. The allele symbols stand for the following: A=round ears, a=pointy ears, B=black fur pigment, b=brown fur pigment, C=long tail, c=short tail, D= whiskers, d=no whiskers, S=spotted, s=solid coat a. How many different phenotypes would be expected among the F2 offpspring? b. what proportion of the F2 offspring will have the round ears, black fur, short tails, whiskers, and a solid coat? c. if 2 F1 mice mate and produce 6 babies, what is the probability they will all have brown fur? d. if 2 F1 mice mate and produce 6 babies, what is the probability that at least one will have black fur?

Solution

Answer:

Each gene will have two allele. A heterozygote of single gene plant produces two types of gametes.

Like wise 2ⁿ; n = nimber of genes

Here 5 genes are there. 2⁵ means 32 gametes will be produced from each heterozygous (all genes) plants

So, 32 * 32 = 1024 genotypes

Number of gemetes = Number of phenotypes

So total numer of phenotypes = 32

b). 27/256

proportion of the round ears (A_), black fur (B_), short tails(cc), whiskers(D_), and a solid coat(ss) = A_B_ccD_ss

=3/4 * 3/4* 1/4 * 3/4* 1/4 = 27/256