Answer:
There is 1/4, or 25% of probabilities of observing an individual with the AayyTtIi genotype in the next generation.
Step-by-step explanation:
Available data:
- For flower position, the axial (A) allele is dominant over terminal (a);
- for seed color, the yellow allele (Y) is dominant over green (y);
- for stem length, the tall (T) allele is dominant over dwarf (t);
- for pod shape, the inflated (I) allele is dominant over constricted (i)
- Test-cross with AAYyTtII genotype
Parental) AAYyTtII x aayyttii
Gametes) AYTI, AyTI, AYtI, AytI
ayti
Punnet Square) AYTI AyTI AYtI AytI
ayti AaYyTtIi AayyTtIi AaYyttIi AayyttIi
There is 1/4, or 25% of probabilities of observing an individual with the AayyTtIi genotype in the next generation.
Note: If you do not want to perform the punnet square for the whole genotypes, you can make crosses for each gene and then multiply the corresponding probabilities to get the desired genotypic ratio. Like this:
Parental) AA x aa
Gametes) A A a a
Punnet square) A A
a Aa Aa
a Aa Aa
There are 4/4 of probabilities of getting Aa genotype in the next generation
Parental) Yy x yy
Gametes) Y y y y
Punnet square) Y y
y Yy yy
y Yy yy
There are 2/4 of probabilities of getting Yy genotype, and 2/4 of probabilities of getting yy genotype in the next generation
Parental) Tt x tt
Gametes) T t t t
Punnet square) T t
t Tt tt
t Tt tt
There are 2/4 of probabilities of getting Tt genotype, and 2/4 of probabilities of getting tt genotype in the next generation
Parental) II x ii
Gametes) I I i i
Punnet square) I I
i Ii Ii
i Ii Ii
There are 4/4 of probabilities of getting Ii genotype in the next generation
The probabilities of observing an individual with the AayyTtIi genotype in the next generation is:
4/4 Aa x 2/4yy x 2/4Tt x 4/4Ii
4/4 x 2/4 x 2/4 x 4/4 = 64/256 = 32/128 = 1/4
There are 1/4 or 25% of probabilities of observing an individual with the AayyTtIi genotype in the next generation