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Do you mean a dihybrid cross? That's the one that has a phenotypic ratio of 9:3:3:1.
From the ratios you listed, both parents had to be heterozygous for all traits. Since you didn't give any alleles or traits, I'll use AaBb to represent the parents' genotypes. So the cross would have been AaBb x AaBb.
The first thing you'd have to do to make this cross is find the possible gametes the parent can produce. One common mistake that students make is to try to put the wrong letters together. Each letter, whether it's upper or lower case represents a specific trait, and each gamete has to include the "instructions" for each of these traits to make a "complete" organism. Remember how we get one of each gene from our mother and one from our father? To find out how many possible gametes you'll get, use the base number 2 (there are two of each allele in a diploid organism, and you can see there are 2 of each letter) and raise it to the power of the number of different traits (letters), in this case 2 (a and b). So you have 2^2, or 2 x 2 (the number 2 multiplied times). In either case, the number for the result is 4. So there are 4 possible gametes.
To find out what the gametes are, use FOIL (the First, Outside, Inside, And Last of each of the pairs of letters:
First - AB
Outside - Ab
Inside - aB
Last - ab
Since there are 4 possible gametes, the Punnett square has to be 4 x 4 blocks, so you have 16 total:
......|..AB..|..Ab..|..aB..|..ab..|
.AB.|
.Ab.|
.aB.|
..ab.|
The next step is to combine the letters from the top of each column with the letters from the right. You can do this yourself by copying what I've given above onto a separate sheet of paper. If you have a letter that's uppercase in either pair, put that first so you can see the dominant genotypes easily. When you're finished, make a list of each UNIQUE genotype, and count how many of each occur. For example, there's only one AABB, but two of AaBB. Once you have the list and the number of each, figure out the phenotype. AABB, AaBB, AABb, and AaBb would all have the same phenotype, since each genotype has at least ONE dominant allele for each trait. If it helps, assign a trait for the letters. Make A = tall and a = short and B = blue and b = white. When you count up the 16 squares, you should find that 9 will be dominant for both traits, 3 will be dominant for A, but recessive for b, 3 will be recessive for a, but dominant for B, and 1 will be recessive for both traits.
If this is a trihybrid cross, the ratio wouldn't be 9:3:3:1 if both parents are heterozygous. There are 64 total possible gametes, and the ratio would be 27:9:9:3:9:3:1. Even if the cross was AaBbCc x aabbcc, the ratio would be different. But here's how to work these. To find out what the gametes are, start will all letters in the uppercase. Then change the last letter every row, the next to the last every second row, the next letter every fourth row. If you have more than 3 traits, always go from right to left, and for each new letter, double the number of rows before you change the case. Keep working till all the letters are in lower case. So for your traits:
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
There are the 8 possible gametes for the parents' genotypes. When you make a Punnett Square, those 8 possibilities means your square will be 8 blocks by 8 blocks:
....|ABC|ABc|AbC|Abc|aBC|aBc|abC|abc|
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
If one parent was aabbcc (AaBbCc x aabbcc), there would only be 1 possible gamete for the second parent (abc).
......|ABC|ABc|AbC|Abc|aBC|aBc|abC|abc|
abc|
although there would technically be 8 abc gametes (the first "a" with first "b" with first "c", first "a" with first "b" with second "c", etc. to second "a" with second "b" with second "c").
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