Dihybrid crosses. Total nightmares, right? AaBb x AAbb, etc. All too often your teachers will show you a giant 4×4 gamete Punnett to find the answer. They’ll tell you that there are too many alleles for a simple Punnett square. But that isn’t true. As a matter of fact, you can use the little squares to solve a problem with any number of genes.
I know, there are a ton of letters. But don’t panic! We won’t need a massive 10×10 diagram. We’re just going to use easy-peasy Punnetts. In this problem, there are 5 genes, so we’re going to need 5 Punnetts. No matter how many genes you have, all you need to do is a Punnett for each gene and then step six, okay? But I’m getting ahead of myself.
Step One: Punnett A
We’re going to ignore every allele in the problem except the “A”s and “a”s. Parent 1 is AA and Parent Two is Aa. So let’s do the Punnett square. If you’ve got those down, skip the italicized section. For those of you still struggling with Punnetts, I’ll do this one in detail: I can’t draw it with this software, so I’ll just describe it. Make a square and split it into fourths. Above the top left, write the 1st A for Parent 1 and place the 2nd A above the top right. To the left of the top left, write the A from Parent Two. To the left of the bottom left, write Parent 2’s a. Then go through and fill each mini square the allele of its column and the allele of its row. The top left mini square will be AA, the top right mini square will be AA, the bottom left mini square will be Aa, and the bottom right mini square will be Aa. That’s 2/4 AA and 2/4 Aa, which simplifies to 1/2 AA and 1/2 Aa. Your Punnett should yield 1/2 AA and 1/2 Aa. Go back to the problem. What “A”s was it asking for in the child? AA, which is a 1/2 probability. But that isn’t your final answer, because we’ve ignored all the other genes so far. But remember 1/2 AA.
Step Two: Punnett B
This time we will only focus on “B”s and “b”s. Parent 1 = BB, Parent 2 = bb. Make the Punnett square just like you did last time. You should get 4/4, or 1/1, Bb. The problem was asking for Bb in the child, so our answer for this gene is 1/1 Bb.
Step Three: Punnett C
Parent 1 = CC, Parent 2 = CC. We don’t even have to make a Punnett for this one because dominant C is the only possible allele. We get 1/1 CC for the child, which is what the question wanted. The child’s gene is 1/1 CC.
Step Four: Punnett D
You know the drill. Parent 1 = Dd and Parent 2 = Dd. The results of the Punnett are 1/4 DD, 1/2 Dd, and 1/4 dd. They’re asking for a child with Dd, so this answer is 1/2 Dd.
Step Five: Punnett E
Parent 1 is EE, Parent 2 is Ee. The Punnett gives us 1/2 EE and 1/2 Ee. For our problem, we’re looking at the probability of a child with EE, so the solution is 1/2 EE.
Step Six: Final Solution
So far, we’ve done five Punnetts. We have the probability of each individual gene, but not the probability of all of them together. So how do we get that? The answer is simple multiplication. Take all of the bolded answers and multiply them together. 1/2 AA x 1/1 Bb x 1/1 CC x 1/2 Dd x 1/2 EE. For the numerator, that’s 1x1x1x1x1 = 1. For the denominator, it’s 2x1x1x2x2 = 8. (2 times 1 equals 2, times 1 again is still 2, times 2 is 4, times 2 is 8). So our answer is that the probability of a child being AABbCCDdEE is 1/8.
About The Author
|English And Biology Ace|
|I am a Keller High School graduate on the Distinguished Achievement Plan. I am matriculating to Brigham Young University in Fall 2017, majoring in Biology (BS). I will be available to tutor in many subjects over the summer. I scored a 31 on the ACT and a 1460 on the SAT. I participated in National H...|