In pea plants flower position, stem length, and seed shape are controlled by ind
ID: 8286 • Letter: I
Question
In pea plants flower position, stem length, and seed shape are controlled by independently assorting genes. Each characteristic has the following mode of inheritance:Character Dominant Recessive
Flower position Axial (A) Terminal (a)
Stem length Tall (T) Dwarf (t)
Seed shape Round (R) Wrinkled (r)
If a plant that is heterozygous for all three characteristics is allowed to self-fertilize, what proportion of the offspring would you expect to be dwarf with terminal flowers and round seeds? Use the probability method to solve this genetic problem.
Explanation / Answer
The heterozygous plant that is self-fertilized has genotype AaTtRr. We want to find the proportion of offspring in the resulting generation with phenotype aattR_ (the last allele for the shape gene can be either R or r in order to have round shape). First we know that the genes independently assort, so we can tackle the probabilities of each gene separately: For flower position we need to get aa from Aa x Aa. The probability of getting each "a" from either parent (Aa) is 1/2, with total probability of 1/2 x 1/2 = 1/4. For stem length, we need to get tt from Tt x Tt. The probability of getting each "t" from either parent (Tt) is 1/2, with total probability of 1/2 x 1/2 = 1/4. For stem length, we need to get either RR or Rr (both represent round shape) from Rr x Rr. For RR, the probability of getting each "R" from either parent (Rr) is 1/2, with total probability of 1/2 x 1/2 = 1/4. For Rr, the probability of getting "R" from one parent and "r" from the other parent is each 1/2, for total probability of 1/2 x 1/2 = 1/4. So the total probability of getting RR OR Rr is 1/4 + 1/4 = 1/2. Now we multiply the total probabilities for each gene to get the probability that an offspring will have aattR_ (i.e. Terminal flowers AND dwarf AND round seeds) = 1/4 x 1/4 x 1/2 = 1/32.