easy way to do dihybrid cross calculations?

[wait4me]

whats the easy way to do dihybrid cross calculations with math??

I know the typical model one LlRr x LlRr is 9:3:3:1, 4 phenotypes 9 genotypes... bc i memorized that fact... but how do you actually do the calculation when both parents arent heterozygotes? what if one trait is codominant?

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[inaccensa]

whats the easy way to do dihybrid cross calculations with math??

I know the typical model one LlRr x LlRr is 9:3:3:1, 4 phenotypes 9 genotypes... bc i memorized that fact... but how do you actually do the calculation when both parents arent heterozygotes? what if one trait is codominant?

think of AABB X AaBb
what gametes will form? AA, Aa,BB,Bb they will form 2/4 X each.
11111111111111--BB (2/4) --AABB 4/16
AA (2/4)------
11111111111111--Bb(2/4)--AABb 4/16


1111111111111--BB(2/4)--AaBB 4/16
Aa (2/4)------
1111111111111--Bb(2/4)--AaBb 4/16

It is easy to figure how many different gametes will form by punnet square and then follow this method to get the actual #. I learned this in advanced genetics. It may seem difficult at first, but trust me once u understand it, you can get the # for anytype of individual.

 

[DJDakoo]

hey...didnt want to make a new thread, but am also looking for ways to do dihybrid crosses without using the table...

I dont really understand the advanced genetics stuff lol...anyone else have other ways? or can explain inaccensa's method?

thanks!

 

[Rabolisk]

In dihybrid crosses, the two genes are independent. So you multiply the percentages, like product rule in math. Let's say we are concerned with a cross between AaBb and Aabb. First treat the A gene. You can easily tell the phenotypic ratio is 3:1 or 3/4 to 1/4. for the other gene, it is 1:1 or 1/2 and 1/2. So you multiply to get the four possible combos. Chance of getting both dominant phenotypes is 3/4 times 1/2 or 3/8. Continuing with this, you get 3/8 for dominant A recessive b, 1/8 for dominant B and recessive a, and 1/8 for both recessive. So the ratio is 3:3:1:1 for this cross. Don't forget that this is for phenotype. Genotypic ratio is calculated the same way, except you also have to differentiate between homozygous and heretozygous. Do not be intimidated by the complexity of the question. Whether it has incomplete dominance, or even more than 2 genes, the method is the same. I hope that made sense!

 

[thomasfx10]

... AaBb and Aabb. First treat the A gene. You can easily tell the phenotypic ratio is 3:1 or 3/4 to 1/4. for the other gene, it is 1:1 or 1/2 and 1/2. So you multiply to get the four possible combos. Chance of getting both dominant phenotypes is 3/4 times 1/2 or 3/8. Continuing with this, you get 3/8 for dominant A recessive b, 1/8 for dominant B and recessive a, and 1/8 for both recessive. So the ratio is 3:3:1:1 for this cross ...

Sorry to resurrect an old thread ... I am trying to find away without spending the precious time of creating a dihybrid cross. Usually when I get a dihybrid question I give it my best guess and move on ... However, I do want to see if there is a quick way without making those pretty tables ...

I searched and found this year old thread ... but I am not understanding this

Looking at AaBb and Aabb. Starting from the A gene ... To get the 3 to 1 are we counting all the "A"'s ... e.g. Aa Aa? then it starts with the other gene stating it is 1:1 then 1/2 ... that is where I am getting lost ... can someone please explain in more detail.

Thanks!

 

[muhali3]

Just separate it into two monohybrid crosses and multiply the probabilities.

For example:

What is the chance of a TtWw child when crossing TTWw and Ttww parents?

1st step: What is the chance for Tt? Do the TT x Tt cross. Answer: 1/2

2nd step: What is the chance for Ww? Do the Ww x ww cross. Answer: 1/2

The answer is the product of the individual probabilities: 1/4.

 

[thomasfx10]

Just separate it into two monohybrid crosses and multiply the probabilities.

For example:

What is the chance of a TtWw child when crossing TTWw and Ttww parents?

1st step: What is the chance for Tt? Do the TT x Tt cross. Answer: 1/2

2nd step: What is the chance for Ww? Do the Ww x ww cross. Answer: 1/2

The answer is the product of the individual probabilities: 1/4.

That seems easy enough ... However I am still doing two Punnet squares

Is there another way to just know that ...

AABB and aabb will yield so and so
or
AaBb and AaBb will be ... so and so


If someone has a quick ref chart please post

 

[salim271]

That seems easy enough ... However I am still doing two Punnet squares

Is there another way to just know that ...

AABB and aabb will yield so and so
or
AaBb and AaBb will be ... so and so


If someone has a quick ref chart please post

Get good enough to do the punnet square in your head? It really doesnt get much easier than that when you're talking about a dihybrid. Especially when you balancing all the different things the genes could mean, dominant, codominant, incomplete dominance, linked, autosomal, mendelian, blah blah blah you get the point. There isnt going to be a lot of hybrid crosses on the MCAT, i cant see anyone getting more than one or two, there are other things about genetics that should and will be tested on.

 

[tn4596]

best way is to make a chart... it doesnt take that long...
at least there is no epistasis in the mcat or else...

 

[notbobtrustme]

I just do the Punnett Square. It's not worth the risk esp when a Punnett square takes like a minute to do, especially if you know what you are looking for (eg # of heterozygotes).

 

[toothhornet88]

So if a question asks to find ratios of the F2 generation and gives you the genotypes of the parents, does that mean I have to first find the ratios of the F1 generation and then use those ratios to find F2?

So if is TTRR x ttrr cross is done F1 will be all TtRr
and to get F2 you need to do a TtRr x TtRr cross to get 9:3:3:1 ratios right?

my question is: if the initial cross is different (say TtRR x ttRR) and you get more than one type of F1 genotype (ttRR, TtRR, TTRr, and ttRr in this case), do you have to make a seperate punnet square cross for each genotype to find the F2 generation?

Thanks in advance!!