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In tetrahybridism with 4 independent genes, the F2 produces:

a) 81 phenotypes for all pairs of alleles in the dominant condition
b) 16 genotypes for all pairs of alleles in the dominant condition
c) 24 phenotypes for one pair of alleles in the codominant condition
d) 24 phenotypes for two pairs of alleles in the codominant condition

i have the right answer, but i don't know how they solved it, so if you know how plz post the method you used to solve it. thx :)

Hey Karim89, duddy gave a detailed explaination to this problem today (go halfway down):

https://biology-forums.com/index.php/topic,98.0.html (https://biology-forums.com/index.php/topic,98.0.html)

ok i understand the explanation, but i can't understand the question asked in here..

here is how i see it ..

a) 81 phenotypes for all pairs of alleles in the dominant condition

so we have a self cross between AaBbCcDd in F1, according to the rule, if all phenotypes in F2 should show only the dominant character, we say that : (3/4)^4= 81/256..

b) 16 genotypes for all pairs of alleles in the dominant condition

so we are looking for those who have the genotype " AABBCCDD " in F2 .. so : (1/4)^4= 1/256

c) d) i don't see how we can have a codominant condition if the genotype of F1 was AaBbCcDd ..


the answer i have says that  " c " is the correct answer... but according what i understood from the previous answer " a " seems the right answer ( if i didn't solve it wrong ) .. so what do you think ?

I see what you're saying.

Well its definitely not (b), but more likely (c). Just the fact that (a) is 81 (an odd number) where you have "4" independent genes (even number) gives it away that its not the answer. Plus, 81 different phenotypes is too high assuming what they mean the offspring is AABBCCDD (a probability of 1/256).

:-[ not the best explanation.

So I think what they are getting at, I'm assuming, is that you will get 24 AaBbCcDd and the probability of this is 1/16 since 0.5 x 0.5  x 0.5  x 0.5 = 1/16 (assuming independent assortment of parents with the alleles AaBbCcDd). I got this by remembering that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4.

we need duddy  :P

haha ;D Probability and genetics was never my favourite subjects. If I think of anything else, I will notify you.

Hey Karim89, I got this explanation from an old journal article.

Note:

In General, an allelomorphic pair is represented by the binomial a + b.
In the binomial "a" = 3 and is dominant, "b" = 1 and is recessive. a : b as 3 : 1.

To detemine the phenotypes:

a) The number of allelomorphic pairs involved gives the power to which the binomial is to be raised. For example, AaBbCc X AaBbCc = (a + b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3.

b) The sum of the coefficients of th expanded binomials gives the number of phenotypes: 1 + 3 + 3 + 1 = 8. So, 8 phenotypes.

c) For any term in the expanded binomial.

   (1) The coefficient indictaes the number of phenotypes,
   (2) 3 raised to the power of the "a" exponent gives the number of individuals in each phenotypes.
   (3) An exponent gives the number of dominants or recessives in each    phenotype.

For examples 3a^2b = (1) 3 phenotypes (2) 9 individuals in each (3) 2 dominants and 1 recessive.

d) Application, assuming 4 allelomorphic pairs (i.e. AaBbCcDd).
(a + b)^4 = a^4 + 4(a^3)b + 6a^2b^2 + 4ab^3 + b^4
1 + 4 + 6 + 4 + 1 = 16 phenotypes.
- a^4 represents 1 phenotypes of 81 (a^4) individuals each having 4 dominants, a^4 = 81 ABCD.
- 4(a^3)b represents 4 phenotypes of 27 (a^3) individuals each, each phenotypes having 3 dominants and 1 recessive,
- 4(a^3)b = 27 ABCd + 27 ABcD + 27 AbCD + 27 aBCD.

Similarly...

6(a^2)(b^2) = 9 ABcd + 9AbCd + 9 AbcD + 9 abCD + 9aBcD + 9 aBCd,
4a(b^3) = 3Abcd + 3aBcd + 3abCd + 3abcD,
b^4 = 1 abcd.

Sorry for the late reply, read the attachment for clarification

The article also explains how to get the genotypes of a particular cross (I didn't write it down here though).

thx this makes a lot of sense, i can't believe we didn't study it this way in class