How do you do dihybrid crosses WITHOUT using punnett squares?

On what basis are you sure there is a better way to get the genotype and phenotype ratios of a dihybrid cross without using a Punnett Square?  If there was an easier/quicker way to do it, then nobody would know what a Punnett Square is.

Here is a website with Punnett Squares and genotype and phenotype ratios for up to trihybrid crosses.

It is sad to see such a close minded response. Of course there are other ways. The forked-line or tree method is what you are after.
This link will show you better than I can explain:



The video is for a trihybrid cross, but still applies in a dihybrid.

Holy sh@#. How do you expect to do a 10 gene cross? By drawing a Punnett square? Open you eyes.

Post Merge: 11 years ago

I second this.

Dihybrid crosses will produce the same types of offspring EVERY single time.
9 A_B_
3 A_bb
3 aaB_
1 aabb

Memorize your ratios for monohybrids then use the above method with multiplying fractions.

I think that Punnett square is the easier way when you observe the transfer of two genes. (So, for the example YyCc X YyCc it would be the best solution.)
The Forked line method is useful mainly for more than two genes. (And again, you don't save too much time)

Different people have different opinions... It doesn't mean that he is close minded because he prefers the punnet square.

I would like to see you making a Forked line method on a 10 gene cross   

You can just multiply the ratios from monohybrids

You know that when 2 heterozygote monohybrids cross, it produces 3:1 phenotypic ratio

Now you have dihybrid A-B-, of which you have 2 alleles, so (3/4)(3/4) would be a 9/16 chance
      -This is assuming you start with heterozygous parents (AaBb)

Then there is trihybrid of which A-B-C-, which has 3 alleles. so (3/4)(3/4)(3/4) would be 27/64
Then there is trihybrid of which aaB-C-, which has 3 alleles. so (1/4)(3/4)(3/4) would be 09/64
   -This would be the same ratio for 1 recessive and 2 dominant phenotypes, doesn't matter if it's A, B, or C which is recessive
Do the same thing for each one you want. ie: 2 recessive and a dominant would yield (1/4)(1/4)(3/4)=3/64

Post Merge: 11 years ago

for your case, YyCc X YyCc

You have a 3/4 chance of Y-dominance, phenotypically
You have a 3/4 chance of C-dominance, phenotypically
So you have a (3/4)(3/4)=9/16 chance of having Y and C phenotypic dominance

Mendel's law of independent assortment says that you have to do each locus separately.

He doesn't "prefer" the punnet square. He is ignorant of any other way to go about it. He then attempts to stifle a member's question by reinforcing his ignorance with circular logic. Attempting to suppress good questions seems pretty close-minded, but I'll keep my mind open.