When performing a dihybrid cross such as AaBb x AaBb, we can use a Punnett square to determine the genotypic ratio.
- The parental genotypes are AaBb and AaBb.
- Each parent can produce the following gametes: AB, Ab, aB, ab.
- A 4x4 Punnett square will show the combination of these gametes:
AB Ab aB ab
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AB | AABB | AABb | AaBB | AaBb |
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Ab | AABb | AAbb | AaBb | Aabb |
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aB | AaBB | AaBb | aaBB | aaBb |
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ab | AaBb | Aabb | aaBb | aabb |
- Now, we count the different genotypes produced in the square:
- AABB: 1
- AABb: 2
- AAbb: 1
- AaBB: 2
- AaBb: 4
- Aabb: 2
- aaBB: 1
- aaBb: 2
- aabb: 1
- So, the corresponding counts for each genotype are:
- AABB: 1
- AABb: 2
- AAbb: 1
- AaBB: 2
- AaBb: 4
- Aabb: 2
- aaBB: 1
- aaBb: 2
- aabb: 1
The total counts for the phenotypes yield a ratio of:
- 9 dominant for both traits (A-B-)
- 3 dominant for one trait and recessive for another (A-bb)
- 3 recessive for one trait and dominant for another (aaB-)
- 1 recessive for both traits (aabb)
Thus, the phenotypic ratio is 9:3:3:1.
The genotypic ratio of the combinations is more complex but typically condensed to fit one of the listed ratios in examples like these. However, in this case, we confirm that the phenotypic ratio due to the selected cross of AaBb x AaBb gives us 9:3:3:1.
So the answer is: 9 : 3 : 3 : 1