Genetics 201 & 202 - Using Punnett Squares for Multiple Genes
Genetics 201: Multi-gene Autosomal
When tracking multiple genes, the biggest difference is that we have to track every combination of genes the parents could possibly donate. But unlike single-gene tracking that we've done so far, multiple mutations can travel separately or together, so we can put some of them into the same parent cell.
To differentiate between sets of autosomal chromosomes, we place a colon between the genes on the Punnett square. For now, I am just going to use color-coding to show how the genes are set up, and how they travel. Here, I have added the parents to the parent chromosome box, both wild type, with 2 pairs of chromosomes shown.
To differentiate between sets of autosomal chromosomes, we place a colon between the genes on the Punnett square. For now, I am just going to use color-coding to show how the genes are set up, and how they travel. Here, I have added the parents to the parent chromosome box, both wild type, with 2 pairs of chromosomes shown.
To populate the parent gene boxes, you need to represent both sides of the colon. Each side contains a set of genes from a pair of chromosomes. Since both parent chromosomes have the same gene, you only have to represent it once, but this won't always be the case. We'll get to that later.
For now, populate the parent gene boxes.
For now, populate the parent gene boxes.
To fill in the offspring cells, we're still taking the genes from the cock row and combining them with the genes from the hen column, but the difference is that each side of the colon is combined separately with the matching side of the color for the other parent bird.
For the first offspring cell, we take the WT from the left side of the cock's colon and combine it with the WT from the left side of the hen's colon, to get WT/WT on the left side of the offspring's colon.
For the first offspring cell, we take the WT from the left side of the cock's colon and combine it with the WT from the left side of the hen's colon, to get WT/WT on the left side of the offspring's colon.
Then we add a colon because we're moving to the next pair of chromosomes, and we take the WT from the right side of the cock's colon and combine it with the WT from the right side of the cock's colon, to get WT/WT on the right side of the offspring's colon. This gives us a WT/WT:WT/WT genotype for the top left offspring cell. This means the offspring is wild type.
Then we do the same thing for all the rest of the cells!
It's a little more complicated than single-gene inheritance, but still looks fairly easy!
I was asked why you can't, for example, take a blue WT from the left side and a brown WT from the right side of the parent hen's genome, and the answer is- you technically can, because they're the same gene, and the color coding is merely there to make it easier to see how to put the genes into the table. It only matters when the genes are not the same, which we'll see in a moment.
I was asked why you can't, for example, take a blue WT from the left side and a brown WT from the right side of the parent hen's genome, and the answer is- you technically can, because they're the same gene, and the color coding is merely there to make it easier to see how to put the genes into the table. It only matters when the genes are not the same, which we'll see in a moment.
Unfortunately the tricky part of multi-gene autosomal Punnett squares comes when dealing with heterozygous birds.
For this example, we will pretend the cock bird is het for 2 different genes, "A" and "B."
Again, I've color coded the parent genes so you can see how to split the parent genes up, and to account for all combos. Because the genes on each side of the colon are NOT all the same within their chromosome pair, we have more possibilities- the cock bird must get 4 gene boxes instead of 2.
For this example, we will pretend the cock bird is het for 2 different genes, "A" and "B."
Again, I've color coded the parent genes so you can see how to split the parent genes up, and to account for all combos. Because the genes on each side of the colon are NOT all the same within their chromosome pair, we have more possibilities- the cock bird must get 4 gene boxes instead of 2.
The cock may donate two wild type genes (1st box), just one A mutation gene (2nd box), just one B mutation gene (3rd box), or both an A and a B mutation gene (4th cell). The hen, as a wild type, will only donate a WT in all slots.
So, again, we just combine the genes from the cock column with the genes in the hen row, careful to match each side of the colon with its same side of the other parent.
So, again, we just combine the genes from the cock column with the genes in the hen row, careful to match each side of the colon with its same side of the other parent.
- The actual order doesn't necessarily matter, as long as all the genes are accounted for and all genes are on the correct side of the colon. I prefer to place the WT designation first in any hets, since it matches saying 'wild type het/split mutation,' which is why some of these are not in the "same" order.
For the above, all the offspring will be blue, visually, but you have actual wild types (WTWT:WTWT), wild type het A (WT/A:WTWT), wild type het B (WTWT:WT/B), and wild type het for both A AND B (WT/A:WT/B). Visually, they all look identical.
In this case, each offspring genotype is still 25% of the total number of offspring, even though there are more than 4 boxes. The first column of offspring all share a genotype, meaning 25% of the total offspring will be full wild types. Same goes for the other columns. This is not always the case. Sometimes every box in an entire Punnett square is different from every other box. You can find the percentage by dividing 100 by the number of offspring boxes (in this case 8, so each box is 12.5%), and multiplying that number by the number of boxes that match that genotype (each genotype has 2 boxes, so 25%).
You can use the above method for any purely-autosomal multi-gene combination. If you have more than two genes to track, all you need to do is add another colon, and account for all gene combos in the parents. Then it's just a matter of filling in boxes. It's not necessarily harder, but it takes longer and there's more room for mistakes the bigger the Punnett square.
In this case, each offspring genotype is still 25% of the total number of offspring, even though there are more than 4 boxes. The first column of offspring all share a genotype, meaning 25% of the total offspring will be full wild types. Same goes for the other columns. This is not always the case. Sometimes every box in an entire Punnett square is different from every other box. You can find the percentage by dividing 100 by the number of offspring boxes (in this case 8, so each box is 12.5%), and multiplying that number by the number of boxes that match that genotype (each genotype has 2 boxes, so 25%).
You can use the above method for any purely-autosomal multi-gene combination. If you have more than two genes to track, all you need to do is add another colon, and account for all gene combos in the parents. Then it's just a matter of filling in boxes. It's not necessarily harder, but it takes longer and there's more room for mistakes the bigger the Punnett square.