Genetics help please?

Question

Four genes are segregating. How many phenotypes are possible among the progeny of a cross between double heterozygotes, if two of the traits are partially dominant (or co-dominat) and two are complete dominants (assume no environmental effect)?

a. 13
b. 4
c. 42
d. 36
e. 8

The answer should be d, 36.

I know what partial dominance is but what is its significance in the problem? How should it be solved? Thank you for your help!

who is smokeyhillbreit???

Both were really great answers I dont know which to choose so I'm going to put it in voting

Answer 1

The two traits that are simple dominant will have phenotypes in a 3 dominant: 1 recessive ratio. Each of these simple dominant traits will have only two possible phenotypes.
(Bb x Bb = BB, Bb, bB, and bb)

The two traits with codominance will have phenotypes in a 1:2:1 ratio. Each of these traits will have three possible phenotypes. (RW x RW = RR, RW, WR, WW)

The number of possible phenotypes when we consider all four of these traits will be 2*2*3*3 = 36.

Answer 2

the significance of the partial dominance is the it affects the phenotypic ratio.

let the alleles be A and a. B and b. C and c. D and d.
let A and a, B and b be in codominance

For A and a you can get either
AA Aa or aa

For B and b you can get either

BB Bb or bb

For C and c you can get either

CC Cc or cc

For D and d you can get either

DD Dd or dd

for A and B there are 3 phenotypes
for C and D there are only 2 phenotypes

so number of possible combinations
(number of possible A)x(number of possible B)x(number of possible C)x(number of possible D)

ie. 3x3x2x2 = 36

Answer 3

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