This tutorial will take you step by step through the question. Individuals with the genotypes Aa Bb CC Dd and Aa bb Cc Dd are crossed. Alleles of these four genes assort independently of one another. Now you can use the step by step method to answer this challenging question. Give the proportion of the progeny of this cross that are expected to have each of the following genotypes. Please answer as a decimal. a) Aa Bb Cc Dd
Number
b) aa bb CC DD
Number
c) Aa bb Cc DD
Number

The correct answer is -

a) 1/16

b) 1/64

c) 1/32

Explanation:

According to the law of independent assortment, each allele of a particular gene has equal chances of getting inherited into the gametes.

In a Mendelian cross of Aa x Aa would give;

AA = 1/4,

Aa = 2/4 and

aa = 1/4

The result of the cross between AaBb x AaBb would be 1/16 due to the fact that probability is still the same AA = 1/4, BB = 1/4 therefore, AABB = 1/16 = 1/4× 1/4.

Now based on this crossing the following:

AaBbCCDd x AabbCcDd.

Take each allele as individual

Aa x Aa would give; AA = 1/4, Aa = 2/4 and aa = 1/4

Bb x bb would give; BB = 1/2, bb = 1/2

CC x Cc would give; CC = 1/2, Cc = 1/2

Dd x Dd would give; DD = 1/4, Dd = 2/4 and dd = 1/4

Then the probability of progeny of the following would be:

a) the probability of AaBbCcDd = 2/4 x 1/2 x 1/2 x 2/4 = 1/16

b) the probability of aabbCCdd = 1/4 x 1/2 x 1/2 x 1/4 = 1/64

c) the probability of AabbCcDD = 2/4 x 1/2 x 1/2 x 1/4 = 1/32

my hotdog is jiggly bruh no kap